In this chapter, we will investigate what have become the standard applications of the integral:

Chapter 8 Overview: Applications of Integrals Calculus, like most mathematical fields, began with trying to solve everyday problems. The theory and operations were formalized later. As early as 70 BC, Archimedes was working on the problem of problem of finding the volume of a non-regular shapes. Beyond his bathtub incident that revealed the relationship between weight volume and displacement, He had actually begun to formalize the limiting process to eplore the volume of a diagonal slice of a cylinder. This is where Calculus can give us some very powerful tools. In geometry, we can find lengths of specific objects like line segments or arcs of circles, while in Calculus we can find the length of any arc that we can represent with an equation. The same is true with area and volume. Geometry is very limited, while Calculus is very open-ended in the problems it can solve. On the net page is an illustration of the difference. In this chapter, we will investigate what have become the standard applications of the integral: Area Volumes of Rotation Volume by Cross-Section Arc Length Just as AP, we will emphasize how the formulas relate to the geometry of the problems and the technological (graphing calculator) solutions rather than the algebraic solutions. 551

Below is an illustration of what we can accomplish with Calculus as opposed to geometry: Geometry Calculus Length Area Volume Calculus can also be used to generate surface areas for odd-shaped solids as well, but that is out of the scope of this class. 55

8.1 Area Between Two Curves We have learned that the area under a curve is associated with an integral. But what about the area between two curves? It turns out that it is a simple proposition, very similar to some problems we encountered in geometry. In geometry, if we wanted to know the area of a shaded region that was composed of multiple figures, like the illustration below, we find the area of the larger and subtract the area of the smaller. The area for this figure would be s ATotal Asquare Acircle s Similarly, since the integral gets us a numeric value for the area between a curve and an ais, if we simply subtract the smaller curve from the larger one and integrate, we can find the area between two curves. y Objectives: Find the area of the region between two curves. 553

Unlike before, when we had to be concerned about positives and negatives from a definite integral messing up our interpretation of area, the subtraction takes care of the negative values for us (if one curve is under the ais, the subtraction makes the negative value of the integral into the positive value of the area. Area Between Two Curves: The area of the region bounded by the curves a and b where f and g are continuous and f b a y f and y g A f g d and the lines g for all in ab, is You can also think of this epression as the top curve minus the bottom curve. If we associate integrals as area under a curve we are finding the area under the top curve, subtracting the area under the bottom curve, and that leaves us with the area between the two curves. The area of the region bounded by the curves f y, g y, and the lines y c and y d where f and g are continuous and f g for all y in cd, is d c A f y g y dy You can also think of this epression as the right curve minus the left curve. Steps to Finding the Area of a Region: 1. Draw a picture of the region.. Sketch a Reimann rectangle if the rectangle is vertical your integral will have a d in it, if your rectangle is horizontal your integral will have a dy in it. 3. Determine an epression representing the length of the rectangle: (top bottom) or (right left). 4. Determine the endpoints the boundaries (points of intersection). 5. Set up an integral containing the limits of integration, which are the numbers found in Step 4, and the integrand, which is the epression found in Step 3. 6. Solve the integral. 554

E 1 Find the area of the region bounded by y sin, y cos, y, and. b A f g d Our pieces are perpendicular to the ais, a so our integrand will contain d b sin cos On our interval a sin cos A d A d Our region etends from A A cos sin cos sin cos sin sin is greater than to cos 555

What if the functions are not defined in terms of but in terms of y instead? y E Find the area of the region bounded by e, y 4, 1 y. 1 y, and y d A c f y g y dy Our pieces are perpendicular to the y ais, so our integrand will contain dy d y A e y 4 dy c 1 y 4 1 y On our interval e is greater than y 4 A e y dy Our region etends from A 4.959 1 y to 1 y 556

E 3 Find the area of the region bounded by y sin, y cos, 0, and. y 4 A d d 0 4 cos sin sin cos.88 With this problem, we had to split it into two integrals, because the top and bottom curves switch partway through the region. We could have also done this with an absolute value: A d 0 cos sin.88 If we are asked this question on an AP test or in a college classroom, they usually want to see the setup. The absolute value is just an easy way to integrate using the calculator. NOTE: Any integral can be solved going dy ord. It is usually the case though, that one is easier than the other. 557

E 4 Sketch the graphs of y 6y and 4y y. If asked to find the area of the region bounded by the two curves which integral would you choose an integral with a dy and an integral with ad? Both will work but one is less time consuming. y We can see that if we went d we would need three integrals (because the tops and bottoms are different on three different sections), but if we go dy we only need one integral. I like integrals as much as the net person, but I ll just do the one integral if that s OK right now. d c 5 A f y g y dy 0 4 6 10y y dy A y y y y dy 5 0 5y 15 3 A y 3 3 5 0 558

8.1 Homework Draw and find the area of the region enclosed by the given curves. 1. y 1, y 9, 1,. y sin, y e, 0, 3. 3 y, y 8, 3, 3 559

4. y sin, y cos, 0, 5. y 5, y 6. 3 y1, y 3 560

7. e y, y, y 1, y 1 8. 1 y, y 1 1 9. y, y, 1, 561

10. y y, y 4y Use your grapher to sketch the regions described below. Find the points of intersection and find the area of the region described region. 11. y, y e, 1 y ln 1, y cos 1. 56

13. y, y 563

Answers: 8.1 Homework 1. y 1, y 9, 1, A=19.5. y sin, y e, 0, A=.810 3. 3 y, y 8, 3, 3 A=60.5 4. y sin, y cos, 0, A=0.5 5. y 5, y A= 3 3 6. y1 3, y 3 A=4 7. y, e y, y 1, y 1 A=5.684 8. 1 y, y 1 A= 8 3 9. y 1, y, 1, A=1.369 10. y y, y 4y A=9 11. y, y e, 1 A=0.350 1. y ln 1, y cos A=1.168 13. y, y A=.106 564

8. Volume by Rotation about an Ais We know how to find the volume of many objects (remember those geometry formulas?) i.e. cubes, spheres, cones, cylinders but what about non-regular shaped object? How do we find the volumes of these solids? Luckily we have Calculus, but the basis of the Calculus formula is actually geometry. If we take a function (like the parabola below) and rotate it around the -ais on an interval, we get a shape that does not look like anything we could solve with geometry. y When we rotate this curve about the ais, we get a shape like the one below: z y 565

This is sort of like a cone, but because the surface curves inward its volume would be less than a cone with the same size base. In fact, we cannot use a simple geometry formula to find the volume. But if we begin by thinking of the curve as having Reimann rectangles, and rotate those rectangles, the problem becomes a little more obvious: y Each of the rotated rectangles becomes a cylinder. The volume of a cylinder is V r h Note that the radius (r) is the distance from the ais to the curve (which is and the height is the change in ( ), giving us a volume of f ) V f 566

To find the total volume, we would simply add up all of the individual volumes on the interval at which we are looking (let s say from = a to = b). total b a V f Of course, this would only give us an approimation of the volume. But if we made the rectangles very narrow, the height of our cylinders changes from to d, and to add up this infinite number of terms with these infinitely thin cylinders, the b becomes b a a giving us the formula total What you may realize is that the b V f d a f is simply the radius of our cylinder, and we usually write the formula with an r instead. The reason for this will become more obvious when we rotate about an ais that is not an - or y-ais. And since integration works whether you have d or dy, this same process works for curves that are rotated around the y-ais and are defined in terms of y as well. Volume by Disc Method (Part 1): The volume of the solid generated when the function f, is rotated about the ais f from = a and = b, where 0 [or g y from y = c and y = d, where g y 0, is rotated about the y ais is given by or b a or V f d V d c V g y dy b r d a where r is the height (or length if they are horizontal) of your Riemann rectangle. 567

Steps to Finding the Volume of a Solid: Objective: 1. Draw a picture of the region to be rotated.. Draw the ais of rotation. 3. Sketch a Reimann rectangle if the piece is vertical your integral will have a d in it, if your piece is horizontal your integral will have a dy in it. Your rectangle should always be sketched perpendicular to the ais of rotation. 4. Determine an epression representing the radius of the rectangle (in these cases, the radius is the function value). 5. Determine the endpoints the region covers. 6. Set up an integral containing outside the integrand, the limits of integration are the numbers found in Step 5, and the integrand is the epression found in Step 4. Find the volume of a solid rotated when a region is rotated about a given line. 568

E 1 Let R be the region bounded by the equations y, the ais,, and 7. Find the volume of the solid generated when R is rotated about the ais. y b r d a V Our pieces are perpendicular to the ais, so our b a integrand will contain d V y d We cannot integrate y with respect to so we will V b d a V 7 d substitute out for y The epression for y is Our region etends from to 7 V 3355 When the region, R, is rotated as in the eample above, the solid generated would look like this: 569

z y Note that knowing that this is what the rotated solid looks like has no bearing on the math. It makes an interesting shape, but being able to draw the image and being able to generate the volume by integration are two completely separate things. We are much more concerned with finding the volume. 570

3 E Let R be the region bounded by the curves y sec, the ais,, 4 5 and. Find the volume of the solid generated when R is rotated about the 4 ais. y b r d a V Our pieces are perpendicular to the ais, so our b y d a integrand will contain d V We cannot integrate y with respect to so we will b substitute out for y V sec d The epression for y is sec a 5 4 V sec d 3 Our region etends from 4 V 3 5 to 4 4 571

Again, when the region, S, is rotated as in the eample above, the solid generated would look like this: z y It s pretty cool-looking, but of no great consequence in terms of the math. 57

What if we had a function defined in terms of y? E 3 Find the volume of the solid obtained by rotating about the y ais the region bounded by y y and the y ais. y d r dy c V Our pieces are perpendicular to the y ais, so our d dy c integrand will contain dy V We cannot integrate with respect to y so we will V d y y dy c V y y dy 0 sub out for The epression for is y y Our region etends from y 0 to y 16 V 15 573

Suppose we wanted to find the volume of a region bounded by two curves that was then rotated about an ais. Volume by Washer Method: The volume of the solid generated when the region bounded by functions f and f g [or g, from = a and = b, where f y and g y, from y = c and y = d, where f y g y ais is given by b a ], is rotated about the or V f g d or d c V f y g y dy b d or V R r d a c V R r dy Where R is the outer radius of your Riemann rectangle and r is the inner radius of your Riemann rectangle. y 574

Again, think of taking a tiny strip of this region, a Riemann rectangle, and rotating it about the ais. Note that this gives us a cylinder (albeit a very narrow one) with another cylinder cut out of it. This is why the formula is what we stated above. If we took all of those strips for the above eample, the solid would look like this. z y Notice that overall, this does not look like a washer per se, but if you cut a cross section, you would see washer shapes like the one illustrated below: Cross section of the solid above: 575

Steps to Finding the Volume of a Solid With the Washer Method: 1. Draw a picture of the region to be rotated.. Draw the ais of rotation. 3. Sketch a Riemann rectangle if the piece is vertical your integral will have a d in it, if your piece is horizontal your integral will have a dy in it. Your rectangle should always be sketched perpendicular to the ais of rotation. 4. Determine an epression representing the radius of the rectangle in the case of washers you will have an epression for the outer radius and an epression for the inner radius. 5. Determine the boundaries the region covers. 6. Set up an integral containing π outside the integrand, the limits of integration are the boundaries found in Step 5, and the integrand is the epression found in Step 4. E 5 Let R be the region bounded by the equations y and y. Find the volume of the solid generated when R is rotated about the ais. y R r 576

b V R r d a Our pieces are perpendicular to the ais, so our integrand will contain d b a V d The epression for R 1 is and the epression 1 0 for R is V d Our region etends from 0 to 1 V.419 E 6 Let R be the region bounded by y, y 1, and 0. Find the volume of the solid generated when R is rotated about the ais. y R r b V R r d a Our pieces are perpendicular to the ais, so our integrand will contain d b 1 a V d The epression for R 1 is 1and the epression 1 1 0 for R is V d Our region etends from 0 to 1 V 1.571 577

E 7 Let R be the region bounded by y 3, 1, and y 0. Find the volume of the solid generated when R is rotated about the y ais. y R r d V R r dy c Our pieces are perpendicular to the y ais, so our integrand will contain dy d c 1 3 V y dy The epression for R 1 is 1and the epression 1 1 3 0 for R is 3 y V y dy Our region etends from y 0 to y 1 V 1.57 578

E 8 Let R be the region bounded by y 1,, and y 10. Find the volume of the solid generated when R is rotated about the y ais. y R r d c y 1 V R r dy 10 5 39.70 dy 579

8. Homework Find the volume of the solid formed by rotating the described region about the given line. 1. y, 1, y 0; about the ais.. y e, 0, 1, y 0; about the -ais. 3. 1 y, 1,, y 0; about the -ais. 4. y 4, y 0; about the -ais. 580

5. y, 0, y 1; about the y-ais. 6. y sec, y 1,, ; about the -ais. 3 3 7. y, y; about the y-ais. 8. y 3, 1, y 0; about the y-ais. 581

9. y 4, y,, ; about the -ais. 10. y, y ; about the -ais. 11. y, y e, 1; about the -ais. 1. y, y ; about the -ais. 58

Answers: 8. Homework 1. y, 1, y 0; about the ais. V= 5. y e, 0, 1, y 0; about the -ais. V=10.036 1 3. y, 1,, y 0; about the -ais. V= 4. y 4, y 0; about the -ais. V= 3 3 5. y, 0, y 1; about the y-ais. V= 5 6. y sec, y 1,, ; about the -ais. 3 3 V=4.303 7. y, y; about the y-ais. V= 64 15 8. y 3, 1, y 0; about the y-ais. V= 3 4 9. y 4, y,, ; about the -ais. V= 59 15 10. y, y ; about the -ais. V= 3 10 11. y, y e, 1; about the -ais. V=1.07 1. y, y ; about the -ais. V=94.61 583

8.3 Volume by Rotation about a Line not an Ais In the last section we learned how to find the volume of a non-regular solid. This section will be more of the same, but with a bit of a twist. Instead of rotating about an ais, let the region be revolved about a line not the origin. Volume by Disc Method (Form ): The volume of the solid generated when the function f, is rotated about the line y k f from = a and = b, where 0 g y from y = c and y = d, where 0 [or given by or b g y, is rotated about the line or V f k d a V d V g y h dy b r d a where r is the Length of your Riemann rectangle. c h is Volume by Washer Method (Part ): The volume of the solid generated when the region bounded by functions f and g, from = a and = b, where f g [or f y and g y, from y = c and y = d, where f y g y is rotated about the line y k is given by b a V f k g k d or d c V f y h g y h dy d V R r dy c where R is the outer radius of the Riemann rectangle and r is the inner radius. ], 584

E 1 Let R be the region bounded by the equations y, y 1, 1, and 1. Find the volume of the solid generated when R is rotated about the line y 1. y r r 1 V b r d a 7 1 3583.333 d As in the previous section, this makes an interesting shape, but is of very little use to us in actually solving for the volume. You could easily have found the value for the volume never knowing what the shape actually looked like. 585

z y 586

3 E Let R be the region bounded by the curves y sec, y,, and 4 5. Find the volume of the solid generated when R is rotated about the 4 line y. y R Rsec b V y d a 5 4 sec 3 4 48.174 d 587

E 3 Find the volume of the solid obtained by rotating about the y ais the region bounded by y y and the line 3 about the line 3. y R 3 R y y d V 3 dy c 3 1 y y dy 3 961.746 What if you were asked to take the region bounded by y and y and rotate it about the ais? How is this different from the previous problems? 588

E 4 Let R be the region bounded by the equations y, the ais,, and 5. Find the volume of the solid generated when R is rotated about the line y. y R r b a 0 V R r d 5 433.478 d 589

3 E 5 Let R be the region bounded by the curves y sec, the ais,, 4 5 and. Find the volume of the solid generated when R is rotated about 4 the line y. b V R r d a 5 4 3 sec 4 d 8.435 y R r The rotated figure looks like this: 590

z y It would be difficult to sketch this by hand, and we definitely do not need it to solve the problem. 591

E 6 Find the volume of the solid obtained by rotating about the y ais the region bounded by y y and the y ais about line 4. Given the information above and the sketch of y y below, sketch your boundaries, the ais of rotation, your Riemann rectangle, and your inner and outer radii. y R r Set up the integration: d V R r dy c 0 4 4 y y dy 30.159 59

E 7 Let R be the region bounded by the curves y sin, y cos, 0, and. Find the volume of the solid obtained by rotating R about the ais. y b a c V R r d R r d b 4 cos sin d sin cos d 0 4 593

8.3 Homework Set A 1. Given the curves f ln, g e and = 4. y a. Find the area of the region bounded by three curves in the first quadrant. b. Find the volume of the solid generated by rotating the region around the line y =. c. Find the volume of the solid generated by rotating the region around the line y = 1. 594

. Let f and g be the functions given by f 41 g 4 1 y and a. Find the area of the region bounded by three curves in the first quadrant. b. Find the volume of the solid generated by rotating the region around the line y = 3. c. Find the volume of the solid generated by rotating the region around the line y =. d. Set up an integral epression for the curve g 4 1 curve h 1 and the which is not pictured above that is rotated around the line y = k, where k >, in which the volume equals 10. Do not solve this equation. 595

Find the volume of the solid formed by rotating the described region about the given line. Sketch the graph so that it is easier for you to apply the Disk/Washer Methods. 3. 1 y, y 0, 1, 3; about the line y 1. 4. y sin, y, 0, ; about the line y 3. 3 5. y, y ; about the line y 1. 3 6. y, y ; about the line 1. 596

Use your grapher to sketch the regions described below. Find the points of intersection and find the volume of the solid formed by rotating the described region about the given line. 7. y ln 1, y cos ; about the line y 1. 8. y, y 4; about the line y 4. 9. y sin, y, 0, ; about the line y. 10. y, 1; about the line 1. 597

8.3 Homework Set B 1 1 g 6 5, find the area 1 bounded by the two curves in the first quadrant. 1. Given the functions f and 3. Find the volume of the solid generated when the function y ln 9 rotated around the line y 4 on the interval 1, is 3. Find the volume of the solid generated when the area in the first quadrant bounded by the curves ln y1 and y ais is rotated around the y 598

4. Let R be the region bounded by the graphs y sin y 1 3 and y 16 4 R a. Find the area of the region R. b. Find the volume of the solid when the region R is rotated around the line y = 10. c. The line y = 3 cuts the region R into two regions. Write and evaluate an integral to find the area of the region below the line. 599

3y 5. Given the functions below are and 4 y. Let A be the region bounded by the two curves and the -ais. y A a. Find the area of region A. b. Find the value of d dy for 4 y at the point where the two curves intersect. c. Find the volume of the solid when region A is rotated around the y- ais. 600

Answers: 8.3 Homework Set A f 1. Given the curves ln, g e and = 4. a. Find the area of the region bounded by three curves in the first quadrant. A=.50 b. Find the volume of the solid generated by rotating the region around the line y =. V=0.54 c. Find the volume of the solid generated by rotating the region around the line y = 1. V=.156. Let f and g be the functions given by f 41 g 4 1 3. and a. Find the area of the region bounded by three curves in the first quadrant. A=1.089 b. Find the volume of the solid generated by rotating the region around the line y = 3. V=19.538 c. Find the volume of the solid generated by rotating the region around the line y =. V=14.689 g 4 1 and the d. Set up an integral epression for the curve curve h 1 which is not pictured above that is rotated around the line y = k, where k >, in which the volume equals 10. Do not solve this equation. 1 10 k g k h d 0 1 y, y 0, 1, 3; about the line y 1. V=8.997 4. y sin, y, 0, ; about the line y 3. V=7.63 601

3 5. y, y ; about the line y 1. V= 5 14 3 6. y, y ; about the line 1. V=1.361 7. y ln 1, y cos ; about the line y 1. V=3.447 8. y, y 4; about the line y 4. V= 51 15 9. y sin, y, 0, ; about the line y. V=9.870 10. y, 1; about the line 1. V= 16 15 8.3 Homework Set B 1 1 bounded by the two curves in the first quadrant. A=4.798 1. Given the functions f and 1 g 6 5, find the area 3. Find the volume of the solid generated when the function y ln 9 rotated around the line y 4 on the interval 1, V=7.86 is 3. Find the volume of the solid generated when the area in the first quadrant bounded by the curves ln y1 and y ais V=14.757 is rotated around the y 60

1 3 4. Let R be the region bounded by the graphs y sin y 16 4 a. Find the area of the region R. R=16 b. Find the volume of the solid when the region R is rotated around the line y = 10. V=766.489 c. The line y = 3 cuts the region R into two regions. Write and evaluate an integral to find the area of the region below the line. 3.55799 1 3 V 16 10 d 5.775 4 0.7796156 and 3y 5. Given the functions below are and 4 y. Let A be the region bounded by the two curves and the -ais. a. Find the area of region A. A=1.609 b. Find the value of d dy for 4 y at the point where the two curves intersect. d 3.578 dy c. Find the volume of the solid when region A is rotated around the y- ais. V=14.986 603

8.4 Volume by the Shell Method There is another method we can use to find the volume of a solid of rotation. With the Disc/Washer method we drew our sample pieces perpendicular to the ais of rotation. Now we will look into calculating volumes when our sample pieces are drawn parallel to the ais of rotation (parallel and shell rhyme that s how you can remember they go together). Objectives: Find the volume of a solid rotated when a region is rotated about a given line. Take the following function, let s call it ais. f, from a to b and rotate it about the y 604

How do we begin to find the volume of this solid? Imagine taking a tiny strip of the function and rotating this little piece around the y ais. What would that piece look like once it was rotated? What is the surface area formula for a cylindrical shell? rl Now would the surface area of just this one piece give you the volume of the entire solid? How would we add up all of the surface areas of all these little pieces? We arrive at our volume formula. Volume by Shell Method: The volume of the solid generated when the region bounded by function g y, from f, from = a and = b, where f 0 [or y = c and y = d, where g y 0], is rotated about the y ais is given by V b f d or V a d c y g y dy ** Everything rhymes with the shell method Shell/Parallel/ rl so the disc method must be of the form Disc/Perpendicular/ r. 605

E 1 Let R be the region bounded by the equations y, y 0, 1, 3. Find the volume of the solid generated when R is rotated about the y ais. y V V b yd Our pieces are parallel to the y ais, so our a integrand will contain d b yd We cannot integrate y with respect to so we will a substitute out for y b d a 3 d 1 V The epression for y is V Our region etends from 1 to 3 V 40 606

E Let R be the region bounded by y, y 5, and 0. Find the volume of the solid generated when R is rotated about the y ais. y b V 5 y d Our pieces are parallel to the y ais, so our a integrand will contain d. We cannot integrate y with respect to so we will substitute out for y. b V 5 d a 5 The epression for y is. V 0 5 d Our region etends from 0 to 5 V 50 4 607

E 3 Let R be the region bounded by y, 1, y and y 4. Find the volume of the solid generated when R is rotated about the ais. y V d c ydy 4 V yy dy 4 3 V 4 y dy V 40 608

E 4 Let R be the region bounded by y, y 3, and 0. Find the volume of the solid generated when R is rotated about the ais. y V d c ydy Our pieces are parallel to the ais, so our integrand will contain dy. But we will need two integrals as our region of rotation is bounded by different curves. The dividing line is y 3 3 3 1 0 3 V y dy y dy 3 3 0 3 V y ydy y y 3 dy V 7 4 The Shell Method can be applied to rotating regions about lines other than the aes, just as the Disk and Washer Methods were. 609

E 5 Let R be the region bounded by the equations y 3, the ais, and 7. Find the volume of the solid generated when R is rotated about the line y 4. y d V c ydy Our pieces are parallel to the ais, so our integrand will contain dy d V c 4 y 7 dy We cannot integrate with respect to y so we will substitute out for d V 4 y 7 y 3 dy The epression for is y 3 c V 4 y 7 y 3 dy Our region etends from y 0 to 0 V 108.909 y 610

The Washer Method yields the same result. b V R r d a b a y1 4 V y d 7 4 4 3 3 V d V 108.909 611

8.4 Homework Set A Find the volume of the solid formed by rotating the described region about the given line. 1. y sec, y 1, 0, ; about the y-ais. 6. y 3, 0, y 8; about the y-ais. 1 y, 1, 8, y 0; about the y-ais. 3. 3 4. y e, y 0,, ; about the -ais. 61

5. y 1 and the -ais; about the y-ais. 6. y sin and the -ais on 0, ; about the y-ais. Use your grapher to sketch the regions described below. Find the points of intersection and find the volume of the solid formed by rotating the described region about the given line. 7. y, y e, 1; about the line 1. 8. y ln 1, y cos ; about the line. 613

9. y, y ; about the line 1. 614

Answers: 8.4 Homework Set A 1. y sec, y 1, 0, ; about the y-ais. 6 V=0.064. y 3, 0, y 8; about the y-ais. V=60.319 1 y, 1, 8, y 0; about the y-ais. V=70.689 3. 3 4. y e, y 0,, ; about the -ais. V=1.969 5. y 1 and the -ais; about the y-ais. V=0.09 6. y sin and the -ais on 0, ; about the y-ais. V= 7. y, y e, 1; about the line 1. V=0.554 8. y ln 1, y cos ; about the line. V=14.676 9. y, y ; about the line 1. V=55.48 615

8.5 Volume by Cross Sections In the last sections, we have learned how to find the volume of a solid created when a region was rotated about a line. In this section, we will find volumes of solids that are not made though revolutions, but cross sections. Volume by Cross Sections Formula: The volume of solid made of cross sections with area A is V b Ad or V a d c A y dy Objectives: Find the volume of a solid with given cross sections. Steps to Finding the Volume of a Solid by Cross Sections: 1. Draw a picture of the region of the base.. Sketch a sample cross section (the base of the sample cross section will lie on the base sketched in Step 1) if the piece is vertical your integral will have a d in it, if your piece is horizontal your integral will have a dy in it. 3. Determine an epression representing the area of the sample cross section (be careful when using semi-circles the radius is half the distance of the base). 4. Determine the endpoints the region covers. 5. Set up an integral containing the limits of integration found in Step 5, and the integrand epression found in Step 4. 616

E 1 The base of solid S is y 1. Cross sections perpendicular to the ais are squares. What is the volume of the solid? z View of the solid generated with the given cross-sections from the perspective of the -ais coming out of the page. y 617

z y View of the solid generated with the given cross-sections from the perspective of the -ais coming out of the page. b a b V A d Start here side V d Area formula for our (square) cross section a 1 V side d Endpoints of the region of our solid 1 1 V side d Simplify the integral 0 1 V y d Length of the side of our cross section V 0 1 4y d We cannot integrate y with respect to so we will 0 1 V 4 1 d 0 V 5.333 sub out for y 618

E The base of solid is the region S which is bounded by the graphs 1 3 y sin and y 16. Cross sections perpendicular to the 4 ais are semicircles. What is the volume of the solid? It is easier for set-up to visualize the base region and the cross-section separately and not to try to imagine the 3D solid. y d d b a b V A d Start here 1 V r d a Area formula for our cross section 4 1 V r d 0 Endpoints of the region of our solid 4 1 d V d 0 1 4 V d d 8 For our cross section y r. 0 4 3 1 V sin 16 8 4 0 d We cannot integrate y with respect to so we will substitute out for y. V 30.08 619

Here is an illustration of the solid, cut-away so that you can see the cross-section: z y d Two views of the solid from Eample z y 60

y 61

E 3 The base of solid is the region S which is bounded by the graphs f 4 1 g 4 1. Cross-sections perpendicular to the and ais are equilateral triangles. What is the volume of the solid? y d V b a 1 0 0 1 3 d d 1 3 d d 0 4 1 3 4 41 1 4 d.589 A d d 6

E 4 Let R be the region in the first quadrant bounded by y cos, y e, and. (a) Find the area of region R. (b) Find the volume of the solid obtained when R is rotated about the line y 1. (c) The region R is the base of s solid. For this solid, each cross section perpendicular to the ais is a square. Find the volume of the solid. (a) cos A e d =.810 0 (b) y R V R r d 0 1 1 1 0 V y y d 1 1 cos e d 0 49.970 63

(c) y side V 0 0 side d cos V e d V 8.045 e cos e cos What you may have noticed is that the base of your cross-section is always either top curve bottom curve or right curve left curve. What this means for us is that all we really need to do is 1) Draw the cross-section and label the base with its length (top bottom or right left). ) Figure out the area formula for that cross-section. 3) Integrate the area formula on the appropriate interval. 64

8.5 Homework Use your grapher to sketch the regions described below. Find the points of intersection and find the volume of the solid that has the described region as its base and the given cross-sections. 1. y, y e, 1; the cross-sections are semi-circles.. y ln 1 and y cos ; the cross-sections are squares. 65

3 3. Given the curves, 4 y quadrant. f 3 4 g and in the first S R T a. Find the area of the regions R, S, and T. b. Find the volume of the solid generated by rotating the curve g around the line y = 8 on the interval 1,3. 4 66

c. Find the volume of the solid generated by rotating the region S around the line y = 1. d. Find the volume of the solid generated if R forms the base of a solid whose cross sections are squares whose bases are perpendicular to the -ais 67

4. Let f and g be the functions given by 3 and g 4 f e y a. Find the area of the region bounded by two curves above. b. Find the volume of the solid generated by rotating the region around the line y = 4. 68

c. Find the volume of the solid generated by rotating the area between g and the -ais around the line y = 0. d. Find the volume of the solid generated if the region above forms the base of a solid whose cross sections are semicircles with bases perpendicular to the -ais. 69

5. Let R and S be the regions bounded by the graphs y 3 y y y and R S a. Find the area of the regions R and S. b. Find the volume of the solid when the region S is rotated around the line = 3. 630

c. Find the volume of the solid when the region R is rotated around the line =. d. Find the volume of the solid if the region S forms the base of a solid whose cross sections are equilateral triangles with bases perpendicular s to the y-ais. (Hint: the area of an equilateral triangle is 3 A ). 4 631

6. Let h be the function given by h ln 1 as graphed below. Find each of the following: a. The volume of the solid generated when h is rotated around the line y on the interval 0, b. The volume of the solid formed if the region between the curve and the lines y 1, 0, and forms the base of a solid whose cross sections are rectangles with heights twice their base and whose bases are perpendicular to the -ais. y 63

c. Find the volume of the solid generated when S is revolved around the horizontal line y d. The region R forms the base of a solid whose cross-sections are rectangles with bases perpendicular to the -ais and heights equal to half the length of the base. Find the volume of the solid. 634

Answers: 8.5 Homework 1. y, y e, 1; the cross-sections are semi-circles. V=0.085 y ln 1 and y cos ; the cross-sections are squares.. V=0.916 3 3. Given the curves, 4 f 3 4 g and in the first quadrant. a. Find the area of the regions R, S, and T. R=.463 S=1.470 T=9.835 b. Find the volume of the solid generated by rotating the curve g around the line y = 8 on the interval 1,3. 4 V=48.15 c. Find the volume of the solid generated by rotating the region S around the line y = 1. V=55.77 d. Find the volume of the solid generated if R forms the base of a solid whose cross sections are squares whose bases are perpendicular to the -ais V=6.96 f e 3 and g 4 4. Let f and g be the functions given by a. Find the area of the region bounded by two curves above. A=7.180 b. Find the volume of the solid generated by rotating the region around the line y = 4. V=599.694 c. Find the volume of the solid generated by rotating the area between g and the -ais around the line y = 0. V=107.33 d. Find the volume of the solid generated if the region above forms the base of a solid whose cross sections are semicircles with bases perpendicular to the -ais. V=66.84 635

5. Let R and S be the regions bounded by the graphs y 3 y and y a. Find the area of the regions R and S. R=.193, S=6.185 b. Find the volume of the solid when the region S is rotated around the line = 3. V=55.345 c. Find the volume of the solid when the region R is rotated around the line =. V=45.081 d. Find the volume of the solid if the region S forms the base of a solid whose cross sections are equilateral triangles with bases perpendicular to the y-ais. (Hint: the area of an equilateral triangle is V=8.607 s 3 A ). 4 6. Let h be the function given by h ln 1 as graphed below. Find each of the following: a. The volume of the solid generated when h is rotated around the line y on the interval 0, V=4.881 b. The volume of the solid formed if the region between the curve and the lines y 1, 0, and forms the base of a solid whose cross sections are rectangles with heights twice their base and whose bases are perpendicular to the -ais. V=1.840 636

1 sin. 4 Let R be the region in the first quadrant enclosed by the y ais and the graphs of f and g, and let S be the region in the first quadrant enclosed by the graphs of f and g, as shown in the figure below. a. Find the area of R R=0.071 b. Find the area of S S=0.38 7. Let f and g be the functions given by f and g e c. Find the volume of the solid generated when S is revolved around the horizontal line y V=5.837 d. The region R forms the base of a solid whose cross-sections are rectangles with bases perpendicular to the -ais and heights equal to half the length of the base. Find the volume of the solid. V=0.007 637

8.6 Arc Length Back in your geometry days you learned how to find the distance between two points on a line. But what if we want to find the distance between to points that lie on a nonlinear curve? Objectives: Find the arc length of a function in Cartesian mode between two points. Just as the area under a curve can be approimated by the sum of rectangles, arc length can be approimated by the sum of ever-smaller hypotenuses. 1 d +dy 1 dy 0.5 0.5 d -1 1-1 1-0.5-0.5 Here is our arc length formula: Arc Length between Two Points: Let f be a differentiable function such that of f over ab, is given by d f ' is continuous, the length L b dy d L 1 d or L 1 dy d dy a c As with the volume problems, we will use Math 9 in almost all cases to calculate arc length. 638

E 1 Find the length of the curve y e 1 from 0 to 3. b L a dy 1 d Start here d L 3 0 1 Substitute in for dy e d d L 19. 58 Math 9 E Find the length of the curve yy 3 from y 1 to y 4. d L 4 c 1 d dy dy L 1 3 y dy 1 L 69. 083 E 3 Use right hand Riemann sums with n = 4 to estimate the arc length of y 3 ln from 1 to 5. How does this approimation compare with the true arc length (use Math 9)? Let f 1 1 ln Let f represent our integrand L 1 1 ln d 1 5 1 f 1 f 3 1 f 4 1 f 5 1. 966. 35. 587. 794 9. 67 Approimate using rectangles 639

Actual Value: 5 L 1 1ln 9. 06 1 Note: The integral is the arc length. Some people get confused as to how rectangle areas can get us length. Remember that the Riemann rectangles are a way of evaluating an integral by approimation. Just like the area under the curve could be a displacement if the curve was velocity, in this case, the area under the curve b a dy 1 d is the length of the arc along the curve y. d 3 E 4 Find the length of the curve f t 1 dt on t 7. L b a b 1 dy d d 3 1 3 1 d a 435. 511 640

8.6 Homework Find the arc length of the curve. 1. 3 y 16 on 0,1 1 1. y ln on, 4 4 1 3 on 1, 9 3 y y y 3. y ln sec on 0, 4 4. 641

5. 3 y ln on 1, 3 6. y e on 0,1 7. Find the perimeter of each of the two regions bounded by y and y. 8. Find the length of the arc along f cos 0 t dt on 0,. 64

9. Find the length of the arc along 4 f 3 t 1 dt on, 1. 643

Answers: 8.6 Homework 1. 3 y 16 on 0,1 L=6.103 1 1. y ln on, 4 4 L=6.499 1 3 on 1, 9 3 y y y L=10.177 3. y ln sec on 0, 4 4. L=0.881 5. 3 y ln on 1, 3 L=1.106 6. y e on 0,1 L=.003 7. Find the perimeter of each of the two regions bounded by y and y. L=34.553 8. Find the length of the arc along f cos 0 L= 9. Find the length of the arc along 4 L=1.337 t dt on 0,. f 3 t 1 dt on, 1. 644

Volume Test 1. The area of the region enclosed by y 4 and y 4 is given by 1 (a) d 0 1 (b) 0 d (c) d 0 (d) 0 4 (e) 0 d d. Which of the following integrals gives the length of the graph y tan between =a to =b if 0 a b? b (a) tan d (b) tan d a b (c) 1sec d (d) 1 tan d a b 4 (e) 1sec d a b a b a 645

3. Let R be the region in the first quadrant bounded by y e, y 1 and ln3. What is the volume of the solid generated when R is rotated about the -ais? (a).80 (b).83 (c).86 (d).89 (e).9 4. The base of a solid is the region enclosed by y cos and the ais for. If each cross-section of the solid perpendicular to the -ais is a square, the volume of the solid is (a) 4 (b) 4 (c) (d) (e) 1 5. A region is bounded by y, the -ais, the line m, and the line m, where m 0. A solid is formed by revolving the region about the - ais. The volume of the solid (a) is independent of m. (b) (c) (d) (e) increases as m increases. decreases as m increases. increases until is none of the above 1 m, then decreases. 646

1 6. Let R be the region in the first quadrant bounded by y sin, the y-ais, and y. Which of the following integrals gives the volume of the solid generated when R is rotated about the y-ais? (a) y dy 0 1 (b) 1 0 sin d sin d (d) sin y 1 (c) (e) sin y 0 1 0 dy 0 dy 647

y 7. Let R be the region bounded by the graphs g 3sin pictured above. 3 (a) Find the area of R. f 1 4 and (b) Find the volume of the figure if R is rotated around the line y = 4. 648

(c) The region, R, is the base of a solid whose cross-sections are squares perpendicular to the -ais. Find the volume of this solid. f 1 4 is rotated around the (d) Find the volume of the solid if only -ais. 649

y 8. Let S be the region bounded by the curves f and 5 (a) Find the area between the two curves. 1 3 g. (b) Find the volume formed by rotating the region S around the line y = 7. 650

Answers: Volume Test 1. The area of the region enclosed by y 4 and y 4 is given by 1 (a) d 0 1 (b) 0 d (c) d 0 (d) 0 4 (e) 0 d d. Which of the following integrals gives the length of the graph y tan between =a to =b if 0 a b? b (a) tan d (b) tan d a b (c) 1sec d (d) 1 tan d a b 4 (e) 1sec d a b a b a 65

y 7. Let R be the region bounded by the graphs f 1 4 and g 3sin pictured above. 3 (a) Find the area of R. R=54.595 (b) Find the volume of the figure if R is rotated around the line y = 4. V=3179.787 (c) The region, R, is the base of a solid whose cross-sections are squares perpendicular to the -ais. Find the volume of this solid. V=549.698 f 1 4 is rotated around the (d) Find the volume of the solid if only -ais. V=1900.035 655

y 1 8. Let S be the region bounded by the curves f 3 and g 5. (a) Find the area between the two curves. S=16.948 (b) Find the volume formed by rotating the region S around the line y = 7. V=430.435 (c) Find the perimeter of the region S. P=7.3 (d) Find the volume formed by rotating the region S around the - ais. V=314.549 656