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Chapter Eight

USING THE DEFINITE INTEGRAL8.1 AREAS AND VOLUMESIn Chapter 5, we calculated areas under graphs using definite integrals. We obtained the integral by slicing upthe region, constructing a Riemann sum, and then taking a limit. In this section, we calculate areas of otherregions, as well as volumes, using definite integrals. To obtain the integral, we again slice up the region andconstruct a Riemann sum.

Finding Areas by SlicingUse horizontal slices to set up a definite integral to calculate the area of the isosceles triangle inFigure 8.1.

Figure 8.1 Isosceles triangle

Example 1

Notice that we can find the area of a triangle without using an integral; we will use this to checkthe result from integration:

To calculate the area using horizontal slices we divide the region into strips; see Figure 8.2. Atypical strip is approximately a rectangle of length w i and width Δh , so

Area of strip≈w iΔh cm 2.

Figure 8.2 Horizontal slicing of isosceles triangle

Solution



Notice that the limits in the definite integral are the limits for the variable h . Once we decide to slice the trianglehorizontally, we know that a typical slice has thickness Δh , so h is the variable in our definite integral, and thelimits must be values of h .

To get w i in terms of h i , the height above the base, use the similar triangles in Figure 8.2:

Summing the areas of the strips gives the Riemann sum approximation:

Taking the limit as n→∞, the change in h shrinks and we get the integral:

Evaluating the integral gives

Use horizontal slices to set up a definite integral representing the area of the semicircle ofradius 7 cm in Figure 8.3.

Figure 8.3 Semicircle

Example 2

As in Example 1, to calculate the area using horizontal slices, we divide the region into strips;see Figure 8.4. A typical strip at height h i above the base has width w i and thickness Δh , so

Area of strip≈w iΔh cm 2.

Figure 8.4 Horizontal slices of semicircle

Solution



Finding Volumes of SlicingTo calculate the volume of a solid using Riemann sums, we chop the solid into slices whose volumes we canestimate.Let's see how we might slice a cone standing with the vertex uppermost. We could divide the cone verticallyinto arch-shaped slices; see Figure 8.5. We could also divide the cone horizontally, giving coin-shaped slices;see Figure 8.6.

Figure 8.5 Cone cut into vertical slices

To get w i in terms of h i , we use the Pythagorean Theorem in Figure 8.4:

so

Summing the areas of the strips gives the Riemann sum approximation

Taking the limit as n→∞, the change in h shrinks and we get the integral:

Using the table of integrals VI-30 and VI-28, or a calculator or computer, gives

As a check, notice that the area of the whole circle of radius 7 is π ·72 = 49π cm 2.



Figure 8.6 Cone cut into horizontal slices

To calculate the volume of the cone, we choose the circular slices because it is easier to estimate the volumes ofthe coin-shaped slices.

Use horizontal slicing to find the volume of the cone in Figure 8.7.

Figure 8.7 Cone

Example 3

Each slice is a circular disk of thickness Δh . See Figure 8.7. The disk at height h i above the

base has radius . From Figure 8.8 and the previous example, we have

w i = 10−2h i so r i = 5−h i .

Figure 8.8 Vertical cross-section of cone in Figure 8.7

Each slice is approximately a cylinder of radius r i and thickness Δh , so

Summing over all slices, we have

Solution



Note that the sum represented by the ∑ sign is over all the strips. To simplify the notation, in the future, we willnot write limits for ∑ or subscripts on the variable, since all we want is the final expression for the definiteintegral. We now calculate the volume of a hemisphere by slicing.

Taking the limit as n→∞, so Δh→0, gives

The integral can be evaluated using the substitution u = 5−h or by multiplying out (5−h )2.Using the substitution, we have

Set up and evaluate an integral giving the volume of the hemisphere of radius 7 cm in Figure8.9.

Figure 8.9 Slicing to find the volume of a hemisphere

Example 4

We will not use the formula for the volume of a sphere. However, our approach can be

used to derive that formula.Divide the hemisphere into horizontal slices of thickness Δh cm . (See Figure 8.9.) Each slice iscircular. Let r be the radius of the slice at height h , so

Volume of slice≈π r 2Δh cm 3.

We express r in terms of h using the Pythagorean Theorem as in Example 2. From Figure 8.10,we have

h 2+r 2 = 72,

so

Thus,

Volume of slice≈π r 2Δh = π (72−h 2)Δh cm 3.

Solution



We now use slicing to find the volume of a pyramid. We do not use the formula, , for the volume of a

pyramid of height h and square base of side length b , but our approach can be used to derive that formula.

Summing the volumes of all slices gives:

Volume≈∑π r 2Δh = ∑π (72−h 2)Δh cm 3.

As the thickness of each slice tends to zero, the sum becomes a definite integral. Since theradius of the hemisphere is 7, we know that h varies from 0 to 7, so these are the limits ofintegration:

Notice that the volume of the hemisphere is half of , as we expected.

Figure 8.10 Vertical cut through center of hemisphere showing relation between r iand h i

Compute the volume, in cubic feet, of the Great Pyramid of Egypt, whose base is a square 755feet by 755 feet and whose height is 410 feet.

Example 5

We slice the pyramid horizontally. creating square slices with thickness Δh . The bottom layeris a square slice 755 feet by 755 feet and volume about (755)2Δh ft 3. As we move up thepyramid, the layers have shorter side lengths. We divide the height into n subintervals of lengthΔh . See Figure 8.11. Let s be the side length of the slice at height h ; then

Volume of slice≈s 2 Δh ft 3.

Figure 8.11 The Great Pyramid

Solution



(a) 1.

Exercises and Problems for Section 8.1Exercises

Write a Riemann sum approximating the area of the region in Figure 8.13, using vertical strips as shown.

Interactive Exploration: Forming a Pyramid with Square Cross-sections

We express s as a function of h using the vertical cross-section in Figure 8.12. By similartriangles, we get

Thus,

and the total volume, V , is approximated by adding the volumes of the n layers:

Figure 8.12 Cross-section relating s and h

As the thickness of each slice tends to zero, the sum becomes a definite integral. Finally, sinceh varies from 0 to 410, the height of the pyramid, we have

Note that , as expected.



(b)

(a)

(b)

2.

(a)

(b)

3.

Figure 8.13

Evaluate the corresponding integral.

Write a Riemann sum approximating the area of the region in Figure 8.14, using vertical strips as shown.

Figure 8.14

Evaluate the corresponding integral.Write a Riemann sum approximating the area of the region in Figure 8.15, using horizontal strips as

shown.

Figure 8.15

Evaluate the corresponding integral.

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(a)

(b)

4.

5.

6.

Write a Riemann sum approximating the area of the region in Figure 8.16, using horizontal strips asshown.

Figure 8.16

Evaluate the corresponding integral.In Exercises 5-12, write a Riemann sum and then a definite integral representing the area of the region, using thestrip shown. Evaluate the integral exactly.




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7.

8.

9.






10.

11.

12.

In Exercises 13-18, write a Riemann sum and then a definite integral representing the volume of the region,using the slice shown. Evaluate the integral exactly. (Regions are parts of cones, cylinders, spheres, andpyramids.)




13.

14.

15.






16.

17.

18.

Problems

The integrals in Problems 19-22 represent the area of either a triangle or part of a circle, and the variable ofintegration measures a distance. In each case, say which shape is represented, and give the radius of the circle orthe base and height of the triangle. Make a sketch to support your answer showing the variable and all otherrelevant quantities.





19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

The integral represents the area of a region between two curves in the plane. Make a sketch

of this region.

In Problems 24-27, construct and evaluate definite integral(s) representing the area of the region described,using:(a) Vertical slices(b) Horizontal slices

Enclosed by y = x 2 and y = 3x .Enclosed by y = 2x and y = 12−x and the y -axis.

Enclosed by y = x 2 and y = 6−x and the x -axis.Enclosed by y = 2x and x = 5 and y = 6 and the x -axis.

The integrals in Problems 28-31 represent the volume of either a hemisphere or a cone, and the variable ofintegration measures a length. In each case, say which shape is represented, and give the radius of thehemisphere or the radius and height of the cone. Make a sketch to support your answer showing the variable andall other relevant quantities.











30.

31.

32. 33.

34.

(a)

(b)

35.

36.

Find the volume of a sphere of radius r by slicing.Set up and evaluate an integral to find the volume of a cone of height 12 m and base radius 3 m.

Find, by slicing, a formula for the volume of a cone of height h and base radius r .Figure 8.17 shows a solid with both rectangular and triangular cross sections.Slice the solid parallel to the triangular faces. Sketch one slice and calculate its volume in terms of x , the

distance of the slice from one end. Then write and evaluate an integral giving the volume of the solid.

Repeat part a for horizontal slices. Instead of x , use h , the distance of a slice from the top.

Figure 8.17

A rectangular lake is 150 km long and 3 km wide. The vertical cross-section through the lake in Figure8.18 shows that the lake is 0.2 km deep at the center. (These are the approximate dimensions of Lake Mead, thelargest reservoir in the US, which provides water to California, Nevada, and Arizona.) Set up and evaluate adefinite integral giving the total volume of water in the lake.









37.

38.

39.

40.

41.

Figure 8.18 Not to scale

A dam has a rectangular base 1400 meters long and 160 meters wide. Its cross-section is shown in Figure8.19. (The Grand Coulee Dam in Washington state is roughly this size.) By slicing horizontally, set up andevaluate a definite integral giving the volume of material used to build this dam.

Figure 8.19 Not to scale

Strengthen Your Understanding

In Problems 38-39, explain what is wrong with the statement.To find the area between the line y = 2x , the y -axis, and the line y = 8 using horizontal slices, evaluate the

integral .

The volume of the sphere of radius 10 centered at the origin is given by the integral .

In Problems 40-41, give an example of:A region in the plane where it is easier to compute the area using horizontal slices than it is with vertical

slices. Sketch the region.A triangular region in the plane for which both horizontal and vertical slices work just as easily.







42.

43.

44.

45.

In Problems 42-45, are the statements true or false? Give an explanation for your answer.

The integral represents the volume of a sphere of radius 3.

The integral gives the volume of a cone of radius r and height h .

The integral gives the volume of a hemisphere of radius r .

A cylinder of radius r and length l is lying on its side. Horizontal slicing tells us that the volume is given by

.

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