1. 7.2 Volumes APPLICATIONS OF INTEGRATION In this section, we will learn about: Using integration to find out the volume of a solid.

2. In trying to find the volume of a solid, we face the same type of problem as in finding areas. We have an intuitive idea of what volume means. However, we must make this idea precise by using calculus to give an exact definition of volume. VOLUMES

3. We start with a simple type of solid called a cylinder or, more precisely, a right cylinder. VOLUMES

4. In general, if the area of the base is A and the height of the cylinder (the distance from B 1 to B 2 ) is h, then the volume V of the cylinder is defined as: V = Ah CYLINDERS

5. In particular, if the base is a circle with radius r, then the cylinder is a circular cylinder with volume V = πr 2 h. CYLINDERS

6. For a solid S that isn’t a cylinder, we first ‘cut’ S into pieces and approximate each piece by a cylinder.  We estimate the volume of S by adding the volumes of the cylinders.  We arrive at the exact volume of S through a limiting process in which the number of pieces becomes large. IRREGULAR SOLIDS

7. For a solid S that isn’t a cylinder, we first ‘cut’ S into pieces and approximate each piece by a cylinder.  We estimate the volume of S by adding the volumes of the cylinders.  We arrive at the exact volume of S through a limiting process in which the number of pieces becomes large. We start by intersecting S with a plane and obtaining a plane region that is called a cross-section of S. IRREGULAR SOLIDS

8. Let A(x) be the area of the cross-section of S in a plane P x perpendicular to the x-axis and passing through the point x, where a ≤ x ≤ b.  Think of slicing S with a knife through x and computing the area of this slice. IRREGULAR SOLIDS

9. We divide S into n ‘slabs’ of equal width ∆x using the planes P x1, P x2,... to slice the solid.  Think of slicing a loaf of bread. IRREGULAR SOLIDS

10. If we choose sample points x i * in [x i - 1, x i ], we can approximate the i th slab S i (the part of S that lies between the planes and ) by a cylinder with base area A(x i *) and ‘height’ ∆x. IRREGULAR SOLIDS

11. So, an approximation to our intuitive conception of the volume of the i th slab S i is: IRREGULAR SOLIDS

12. Adding the volumes of these slabs, we get an approximation to the total volume (that is, what we think of intuitively as the volume):  This approximation appears to become better and better as n → ∞.  Think of the slices as becoming thinner and thinner. Therefore, we define the volume as the limit of these sums as n → ∞). We recognize the limit of Riemann sums as a definite integral and so we have the following definition. IRREGULAR SOLIDS

13. Let S be a solid that lies between x = a and x = b. If the cross-sectional area of S in the plane P x, through x and perpendicular to the x-axis, is A(x), where A is a continuous function, then the volume of S is: DEFINITION OF VOLUME. DISK METHOD.

14. Notice that, for a cylinder, the cross-sectional area is constant: A(x) = A for all x.  So, our definition of volume gives:  This agrees with the formula V = Ah. VOLUMES

15. Show that the volume of a sphere of radius r is Example 1 SPHERES

16. If we place the sphere so that its center is at the origin, then the plane P x intersects the sphere in a circle whose radius, from the Pythagorean Theorem, is: Example 1 SPHERES

17. So, the cross-sectional area is: Example 1 SPHERES

18. Using the definition of volume with a = -r and b = r, we have: (The integrand is even.) Example 1 SPHERES

19. The figure illustrates the definition of volume when the solid is a sphere with radius r = 1. Notice that as we increase the number of approximating cylinders, the corresponding Riemann sums become closer to the true volume. SPHERES

20. Find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to 1. Illustrate the definition of volume by sketching a typical approximating cylinder. Example 2 VOLUMES

21. The region is shown in the first figure. If we rotate about the x-axis, we get the solid shown in the next figure.  When we slice through the point x, we get a disk with radius. VOLUMES Example 2

22. The area of the cross-section is: The volume of the approximating cylinder (a disk with thickness ∆x) is: Example 2 VOLUMES

23. The solid lies between x = 0 and x = 1. So, its volume is: VOLUMES Example 2

24. Find the volume of the solid obtained by rotating the region bounded by y = x 3, y = 8, and x = 0 about the y-axis. Example 3 VOLUMES

25. As the region is rotated about the y-axis, it makes sense to slice the solid perpendicular to the y-axis and thus to integrate with respect to y.  Slicing at height y, we get a circular disk with radius x, where VOLUMES Example 3

26. So, the area of a cross-section through y is: The volume of the approximating cylinder is: Example 3 VOLUMES

27. Since the solid lies between y = 0 and y = 8, its volume is: Example 3 VOLUMES

28. The region R enclosed by the curves y = x and y = x 2 is rotated about the x-axis. Find the volume of the resulting solid. Example 4 VOLUMES – WASHERS METHOD

29. The curves y = x and y = x 2 intersect at the points (0, 0) and (1, 1).  The region between them, the solid of rotation, and cross-section perpendicular to the x-axis are shown. VOLUMES Example 4

30. A cross-section in the plane P x has the shape of a washer (an annular ring) with inner radius x 2 and outer radius x. Example 4 VOLUMES

31. Thus, we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle: VOLUMES Example 4

32. Thus, we have: Example 4 VOLUMES

33. Find the volume of the solid obtained by rotating the region in Example 4 about the line y = 2. Example 5 VOLUMES

34. Again, the cross-section is a washer. This time, though, the inner radius is 2 – x and the outer radius is 2 – x 2. VOLUMES Example 5

35. The cross-sectional area is: Example 5 VOLUMES

36. So, the volume is: VOLUMES Example 5

37. The solids in Examples 1–5 are all called solids of revolution because they are obtained by revolving a region about a line. SOLIDS OF REVOLUTION

38. In general, we calculate the volume of a solid of revolution by using the basic defining formula SOLIDS OF REVOLUTION

39. We find the cross-sectional area A(x) or A(y) in one of the following two ways. SOLIDS OF REVOLUTION

40. If the cross-section is a disk, we find the radius of the disk (in terms of x or y) and use: A = π(radius) 2 WAY 1

41. If the cross-section is a washer, we first find the inner radius r in and outer radius r out from a sketch.  Then, we subtract the area of the inner disk from the area of the outer disk to obtain: A = π(outer radius) 2 – π(outer radius) 2 WAY 2

42. Find the volume of the solid obtained by rotating the region in Example 4 about the line x = -1. Example 6 SOLIDS OF REVOLUTION

43. The figure shows the horizontal cross-section. It is a washer with inner radius 1 + y and outer radius Example 6 SOLIDS OF REVOLUTION

44. So, the cross-sectional area is: Example 6 SOLIDS OF REVOLUTION

45. The volume is: Example 6 SOLIDS OF REVOLUTION