Math Dept, Howard...

Introduction to Volumes using IntegrationExample 1: Volume of Pyramid

Volume of Revolution

1-16-2020 Notes, Calculus 2Introduction to Calcullating Volume

Sankar Sitaraman – nature-lover.net/math

Math Dept, Howard University

1-16-2020

Sankar Sitaraman – nature-lover.net/math 1-16-2020 Notes, Calculus 2



Outline

1 Introduction to Volumes using Integration

2 Example 1: Volume of PyramidVolume of frustum of pyramid

3 Volume of Revolution




Main Ideas

FINDING VOLUMES OF SOLIDSBreak down into slabs whose volumes can be easily found

Solids whose volumes we know how to compute,“without” calculus:

Cylindrical slab of thickness t has volume πr2t .Rectangular slab of length l , width w and thickness t hasvolume lwt .

KEY ITEMS TO LOOK FORBetter to use “vertical” or “horizontal” slabs?



Volume of RevolutionVolume of frustum of pyramid

Volume of Pyramid of square base

Let a be length of base and h the height of pyramid.

We will add volumes of slabs of thickness ∆y and let ∆y → 0.

The width of the slabs decreases as y goes from 0 to h.

Can you tell the width at a height of y , using similar triangles?see picture below




Volume of Pyramid of square base – picture

Figure I shows the triangular cross-section of a pyramid withsquare base. We need to find the length marked ? using similartriangles.




Volume of Pyramid of square base – answer

At a height of y , the width ish − y

ha.

So volume of slab of thickness ∆y located at height y is(h − y

ha)2

∆y .

Adding up these volumes and taking the limit is symbolicallywritten as:∫ h

0

(h − y

ha)2

dy =a2

h2

∫ h

0(h−y)2 dy =

a2

h2

∫ h

0(h2−2hy+y2) dy

=a2

h2

[h2y − 2h(y2/2) + (y3/3)

]h

0=

a2

h2 (h3−h3+(h3/3)) = a2h/3.




Volume of frustum

The volume of a frustum (of height H, base of length a, top oflength b ) can be found by :Volume of frustum = Volume of full pyramid minus Volume of"missing" pyramid.Can you find it using integration? See Figure II below forfrustum.




Volume of frustum – set up

We can use the same integral as before, because the idea isthe same: Add up the volumes of square slabs until you reacha height of h.The only thing that changes is that the height of the full pyramidis not h.Let this height be H. Then using similar triangles as before, we

getH − h

H=

ba.

H − hH

=ba

=⇒ HH− h

H=

ba

=⇒ 1− hH

=ba

=⇒ hH

= 1− ba.

Solving, we get H =ha

a− b.




Volume of frustum – integration

Now we are ready to integrate. All we need to do is to replacethe h inside the integral with H.The limits of the integral are still 0 and h because h is theheight of the frustum.∫ h

0

(H − y

Ha)2

dy =a2

H2

∫ h

0(H − y)2 dy

Note that the integration can be also done using thesubstitution u = H − y and that’s what we do.




Volume of frustum – integration -2

du = −dy =⇒ dy = −du and wheny = 0 =⇒ u = H, y = h =⇒ u = H − h. Integral becomes

a2

H2

∫ H−h

Hu2(−du) = − a2

H2

[u3

3

]H−h

H= − a2

3H2

((H − h)3 − H3

).

Now put H = ha/(a− b) and check that this givesVolume = h(a2 + ab + b2)/3 as mentioned before.




Volume of frustum of cone – set up

Problem 6.2.48: Find volume of frustum of right circular conewith height h, lower base radius R and upper base radius r .

1 Draw the picture.2 Decide whether you want to integrate over x or y , and the

limits of integration.3 Think about how you are going to slice the cone into disks

of thickness ∆y or ∆x .4 Put together everything and write the integral.




Volume of frustum of cone – picture

R is radius of base of conical frustum. r is radius of top.A disk (slice) is shown. Actually it won’t be a proper disk, butsince we take the limit as ∆y → 0 it is okay. What would be theradius r(y) of this disk at height y? What would be the volumeof the disk? Can you write the integral now?




Volume of frustum of cone – integral

Using similar triangles as before will be a bit cumbersome,since we don’t have the full cone. Instead, let us try to write theequation of the line l . Imagine the origin (0,0) at the center ofthe base. Then the line l passes through (R,0) and (r ,h). Its

equation is y − 0 =h

r − R(x − R). Solving for x , we get

x = r(y) = R +(r − R)y

h.

Now set up the integral using the formula for volume of disk.




Volume of frustum of cone – integral -2

The integral is∫ h

0π

(R +

(r − R)yh

)2

dy =π

h2

∫ h

0(Rh + (r − R)y)2 dy

=π

h2

[(Rh + (r − R)y)3

3(r − R)

]h

0=π((rh)3 − (Rh)3)

3(r − R)h2 =π

3h(R2+Rr+r2).

Notice how this is similar to formula for volume of frustum ofPyramid.What happens when r = 0 ? In that case we get the full coneand you can see by plugging in r = 0 in the formula at the endthat the volume of the full cone is πhR2/3.




VOLUME OF SOLID OF REVOLUTION

The idea is the same as for finding the volume of pyramidor cone. Only difference is that we need to visualize thesolid first. The dimensions and boundaries of the solid arenot given to us.

Key Point: When using disks or washers of solids of rev-olution,slice perpendicular to axis of rotation and inte-grate along axis of rotation

PROBLEM 5, 6.2: Rotate region betweenx = 2

√y (y = x2/4), x = 0, y = 9 about the y−axis.

First graph it. What radius (in terms of y ) would a cylindricalslice at height y and thickness ∆y have? Can you write downits volume?




Problem 6.2.5 graph

The parabola is rotated about its axis (y -axis).




Problem 6.2.5 integral

If you slice it at a height of y then you get a cylindrical disk ofradius x and volume πx2∆y = π(4y)∆y . This is because

radius of disk is r(y) = x itself and ro πr2 = πx2 and then wesolve for x2 from the equation y = x2/4.The volume is given by∫ 9

04πy dy = 4π

∫ 9

0y dy = 4π[92/2] = 162π.




Problem 6.2.8

Find the volume of solid obtained by rotating the regionbetween y = 6− x2 and y = 2 about the x-axis.

First graph it.When you integrate along x what radius (in terms of x) would ahollow slice (washer) at x and thickness ∆x have?What are the limits of integration?Can you write down its volume?

Do the same for rotating about y -axis and integration over y .




Problem 6.2.8 graph

The parabola is rotated about x-axis. Resulting solid whensliced gives a washer shaped object. When rotated abouty -axis get an inverted bowl.

outer radius of washerInner radius

of washer




Problem 6.2.8 rotated about y axis

When rotated about y -axis get an inverted bowl.

Volume of disk at height y is πx2dy = π(6− y)dy . This isbecause radius of disk is r(y) = x itself and ro πr2 = πx2 andthen we solve for x2 from the equation y = 6− x2.Volume is given by the integral

∫ 6

2π(6− y)dy = π

[6y − y2

2

]6

2= π(24− 16) = 8π units.




Problem 6.2.8 rotated about x axis

Volume of washer at x is π(R2 − r2)dx where R is outer radiusand r is inner radius.This comes toπ(y2 − 22)dx = π((6− x2)2 − 4)dx = π(32 + x4 − 12x2)dx .Volume is given by the integral∫ 2

−2π(32 + x4 − 12x2)dx = π

[32x +

x5

5− 4x3

]2

−2

= π(64 + (64/5)) = 384π/5 units.


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