Calculus II. Computing the volume of solids.

February 4, 2015 1 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region.

You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is

` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is

` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is

` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is

` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is

` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is

` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is

` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is

` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is` · h · w

2 The volume of a circular cylinder with height h and radius r is

(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is` · h · w

2 The volume of a circular cylinder with height h and radius r is(base)·(height) =

πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is` · h · w

2 The volume of a circular cylinder with height h and radius r is(base)·(height) = πr2h

3 The volume of a ball with radius r is

43πr

3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Cross sections

Using Calculus you are pretty good at computing thearea of a region. You compute it as the integral ofthe length of a cross section.

You probably know the volume of some 3-dimensional solids:

1 The volume of a rectangular box with sides of length `, h, and w is` · h · w

2 The volume of a circular cylinder with height h and radius r is(base)·(height) = πr2h

3 The volume of a ball with radius r is 43πr3

Can we use the techniques of calculus to computevolumes of complicated solids?For example, this one:

February 4, 2015 2 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.

Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)

All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)

All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].

Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].

Cut the solid with a vertical plane at some fixed x-value, x1.

a b

x1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a b

x1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1

This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1

This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.

Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)

today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)

dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Using cross sections to approximate the volume

We hope to compute the volume of this solid.Line it up so that it is parallel to an axis. (I’ll use thex-axis.)All of the x-values that appear in the solid live in aninterval [a, b].Cut the solid with a vertical plane at some fixed x-value, x1. a bx1This two dimensional region is a cross section at x1.Let A(x1) denote the area of this cross section. (The cross sectional area)today’s main result: The volume of the solid is given by integrating thecross sectional area:

Volume =

∫ ba

A(x)dx

Let’s finish the argument about why this is true. It will be reminiscent ofwhy Riemann sums give the area under the curve.

February 4, 2015 3 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx

Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1

a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1

a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .

In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1

a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .

In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x . a bx1

a b

x1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .

At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x . a bx1

a b

x1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .

At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x . a bx1

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.

Build a bunch of cylinders with base given by thecross sections and height by ∆x . a bx1

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.

Build a bunch of cylinders with base given by thecross sections and height by ∆x . a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which

(1) the volume is easy to computeand (2)

well approximate the volume of the original solid.

What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2)

well approximate the volume of the original solid.What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2) well approximate the volume of the original solid.What are the volume of these cylinders?

volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2) well approximate the volume of the original solid.What are the volume of these cylinders?volume of a cylinder = (base)·(height) =

A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2) well approximate the volume of the original solid.What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .

If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2) well approximate the volume of the original solid.What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:

Volume of the original solid ∼∑i

A(xi )∆x

February 4, 2015 4 / 14

Getting an integral

The volume of the solid is given by

∫ ba

A(x)dx Why?

Step 1.Break the interval [a, b] into a bunch of pieces eachof length ∆x .In each of these intervals, pick some xi .At each xi draw the cross section.Build a bunch of cylinders with base given by thecross sections and height by ∆x .

a bx1a b

x1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7

a bx1 x2 x3 x4 x5 x6 x7The point of this construction is replace a solid that is hard to understandwith lots of solids (cylinders) for which (1) the volume is easy to computeand (2) well approximate the volume of the original solid.What are the volume of these cylinders?volume of a cylinder = (base)·(height) =A(xi ) ·∆x .If you add them up you get an approximation of the volume of the originalsolid:Volume of the original solid ∼

∑i

A(xi )∆xFebruary 4, 2015 4 / 14

finishing the argument that integration gives volume

So, given any solid, if A(x) is the area of the cross section at x , and youtake some subdivision of [a, b] into intervals of length ∆x

then the volume of the original solid ∼∑i

A(xi )∆x .

If you make ∆x go to zero then the error in the estimation goes to zero:

Volume of the original solid = lim∑i

A(xi )∆x .

This is an integral!

Volume of the original solid =

∫ ba

A(x)dx .

On the next slide we’ll make use of this

February 4, 2015 5 / 14

finishing the argument that integration gives volume

So, given any solid, if A(x) is the area of the cross section at x , and youtake some subdivision of [a, b] into intervals of length ∆x

then the volume of the original solid ∼∑i

A(xi )∆x .

If you make ∆x go to zero then the error in the estimation goes to zero:

Volume of the original solid = lim∑i

A(xi )∆x .

This is an integral!

Volume of the original solid =

∫ ba

A(x)dx .

On the next slide we’ll make use of this

February 4, 2015 5 / 14

finishing the argument that integration gives volume

So, given any solid, if A(x) is the area of the cross section at x , and youtake some subdivision of [a, b] into intervals of length ∆x

then the volume of the original solid ∼∑i

A(xi )∆x .

If you make ∆x go to zero then the error in the estimation goes to zero:

Volume of the original solid = lim∑i

A(xi )∆x .

This is an integral!

Volume of the original solid =

∫ ba

A(x)dx .

On the next slide we’ll make use of this

February 4, 2015 5 / 14

finishing the argument that integration gives volume

So, given any solid, if A(x) is the area of the cross section at x , and youtake some subdivision of [a, b] into intervals of length ∆x

then the volume of the original solid ∼∑i

A(xi )∆x .

If you make ∆x go to zero then the error in the estimation goes to zero:

Volume of the original solid = lim∑i

A(xi )∆x .

This is an integral!

Volume of the original solid =

∫ ba

A(x)dx .

On the next slide we’ll make use of this

February 4, 2015 5 / 14

finishing the argument that integration gives volume

So, given any solid, if A(x) is the area of the cross section at x , and youtake some subdivision of [a, b] into intervals of length ∆x

then the volume of the original solid ∼∑i

A(xi )∆x .

If you make ∆x go to zero then the error in the estimation goes to zero:

Volume of the original solid = lim∑i

A(xi )∆x .

This is an integral!

Volume of the original solid =

∫ ba

A(x)dx .

On the next slide we’ll make use of this

February 4, 2015 5 / 14

finishing the argument that integration gives volume

So, given any solid, if A(x) is the area of the cross section at x , and youtake some subdivision of [a, b] into intervals of length ∆x

then the volume of the original solid ∼∑i

A(xi )∆x .

If you make ∆x go to zero then the error in the estimation goes to zero:

Volume of the original solid = lim∑i

A(xi )∆x .

This is an integral!

Volume of the original solid =

∫ ba

A(x)dx .

On the next slide we’ll make use of this

February 4, 2015 5 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.

Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section

a circle of radiusr(x) =

x

.

What is the area of this cross section? A(x) =

πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.

If you start at the top of the cone and travel down adistance x you get as a cross section

a circle of radiusr(x) =

x

.

What is the area of this cross section? A(x) =

πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section

a circle of radiusr(x) =

x

.

What is the area of this cross section? A(x) =

πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section

a circle of radiusr(x) =

x

.What is the area of this cross section? A(x) =

πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section

a circle of radiusr(x) =

x

.What is the area of this cross section? A(x) =

πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) =

x

.

What is the area of this cross section? A(x) =

πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .

What is the area of this cross section? A(x) =

πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) =

πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) = πr(x)2 =

πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) = πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) = πr(x)2 = πx2

What interval does h live in?

[0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) = πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) = πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) = πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) = πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

Computing the volume of a cone

Theorem

Let S be a solid sitting in space. For any given x-value let A(x) be thearea of the cross section of S at x. If the x-values of S all live in the

interval [a, b] then the volume of S is

∫ ba

A(x)dx.

example: The volume of a cone.Consider a cone of radius 2 and height 2.If you start at the top of the cone and travel down adistance x you get as a cross section a circle of radiusr(x) = x .What is the area of this cross section? A(x) = πr(x)2 = πx2

What interval does h live in? [0, 2].

Integrate to get the volume: V (cone) =

∫ 20πx2dx =

8π

3.

Does this agree with a formula you know for the volume of a cone?

February 4, 2015 6 / 14

For you

Compute the volume of a cone of radius 1 and height 3.

1 Start at the top of the cone and travel down a distance of x . Drawthe cross section.

2 What is the radius r(x)? Hint: r(0) = 0, r(3) = 1 and r is linear.

3 What is the cross sectional area A(x)?

4 What interval does x live in?

5 Set up and compute the integral.

February 4, 2015 7 / 14

Plan of attack

In order to compute the volume of a solid:

1 Find an axis (x, y or z) so that when you slice perpendicular to thataxis you get a region whose area is easy (or just possible) to compute.

2 What x-values (or y or z if you use those axes) do you see in thesolid? This gives the bounds of integration

3 Compute the cross sectional area A(x) (or A(y) or A(z))

4 Integrate to find the volume.

Notice that steps 2 and 3 can be done in either order.

February 4, 2015 8 / 14

Volume of a sphere

Lets do the same for the unit sphere centered at the origin (0, 0, 0).

1 Since the sphere is so symmetric it does’t matter what axis we use.Let’s use the x axis.

2 In what interval does the x-value of a point in the sphere live? Itmight be helpful to draw a picture. (This gives the bounds ofintegration.)

3 Fix an x-value. Describe the cross section you get. What is the crosssectional area A(x)? (Think about the equality x2 + y2 + z2 = 1 thatdefines the sphere.)

4 Compute the integral, What do you get?

February 4, 2015 9 / 14

Revolving the region under a curve.

Here is the region under the curve y =√x where x runs from 0 to 4.

Imagine revolving this region about the x-axis.Let’s compute the volume.What range of values can x take?Draw an x-cross section. Describe it.What is A(x)?Express the volume as an integral and computeit.

February 4, 2015 10 / 14

Revolving the region between curves

Here is the region between the curve y =√x and y = x (what range of

x-values?)

.Imagine revolving this region about the x-axis.Let’s compute the volume.What range of values can x take?Draw an x-cross section. Describe it.What is A(x)?Express the volume as an integral and computeit.

February 4, 2015 11 / 14

Revolving the region between curves: The method ofwashers look at the addendumIn general if you rotate about the x-axis the region bounded below byy = f (x) and above by g(x) where x runs from a to b and need thevolume, you must compute

volume =

∫ baπ · (g(x)2 − f (x)2)dx .

Perhaps more illuminating notation is

volume =

∫ baπ · (rout(x)2 − rin(x)2)dx .

where rout(x) is the outer radius of the washer at x , and rin(x) is theinner radius of the washer at xDraw the region bounded above by y = mx and below by y = 0 where xruns from 0 to b.Draw the solid given by rotating this region about the x-axis. Describe it.Compute its volume.

February 4, 2015 12 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the regionWhat shape do you get when you rotate it?A Washer!

What is the radius of the inner circle of the washer?

. . . . 1− x

What is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the region

What shape do you get when you rotate it?A Washer!

What is the radius of the inner circle of the washer?

. . . . 1− x

What is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the region

What shape do you get when you rotate it?A Washer!

What is the radius of the inner circle of the washer?

. . . . 1− x

What is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the regionWhat shape do you get when you rotate it?

A Washer!

What is the radius of the inner circle of the washer?

. . . . 1− x

What is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the regionWhat shape do you get when you rotate it?

A Washer!

What is the radius of the inner circle of the washer?

. . . . 1− x

What is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the regionWhat shape do you get when you rotate it?A Washer!

What is the radius of the inner circle of the washer?

. . . . 1− x

What is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the regionWhat shape do you get when you rotate it?A Washer!

What is the radius of the inner circle of the washer?

. . . . 1− x

What is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the regionWhat shape do you get when you rotate it?A Washer!

What is the radius of the inner circle of the washer? . . . .

1− x

What is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the regionWhat shape do you get when you rotate it?A Washer!

What is the radius of the inner circle of the washer? . . . . 1− xWhat is the radius of the outer circle of the washer?

1

The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

Example

Suppose that you want to rotate the region bounded by y = x , y = 0 andx = 1 about the line y = 1.

Since we are rotating about a horizontal line, we willconsider a generic vertical line in the regionWhat shape do you get when you rotate it?A Washer!

What is the radius of the inner circle of the washer? . . . . 1− xWhat is the radius of the outer circle of the washer? 1The volume of this solid of revolution is given by∫ baπ(r2out − r2in) =

∫ 10π(12 − (1− x)2)dx .

February 4, 2015 13 / 14

For you

Here is the region under the curve y =√x (OR x = y2) where x runs

from 0 to 4.Imagine revolving this region about the y -axis. Sketch a picture.Let’s compute the volume.Draw a generic horizontal cross section at some y -value (since the axis ofrevolution is vertical) What are its endpoints in terms of y? What is thedistance from each of the endpoints to the axis of revolution?rin(y) =rout(y) =What is the y -interval of interest?

Compute the integral

∫ baπ(rout(y)

2 − rin(y)2)dx .

February 4, 2015 14 / 14