1 / 34

SECTION 7.2

SECTION 7.2. VOLUMES. DEFINITION OF VOLUME. 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:.

ezra-flores
Download Presentation

SECTION 7.2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. SECTION 7.2 VOLUMES

  2. DEFINITION OF VOLUME Let S be a solid that lies between x = a and x = b. If the cross-sectional area of S in the plane Px, through x and perpendicular to the x-axis, is A(x), where A is a continuous function, then the volume of S is: 7.2

  3. Example 1 • Show that the volume of a sphere of radius r is 7.2

  4. Example 1 SOLUTION • If we place the sphere so that its center is at the origin, then the plane Px intersects the sphere in a circle whose radius, from the Pythagorean Theorem, is: • So, the cross-sectional area is: 7.2

  5. Example 1 SOLUTION • Using the definition of volume with a =– r and b = r, we have: 7.2

  6. Example 2 • 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. 7.2

  7. Example 2 SOLUTION • The region is shown in Figure 6(a). If we rotate about the x-axis, we get the solid shown in Figure 6(b). • When we slice through the point x, we get a disk with radius . 7.2

  8. Example 2 SOLUTION • The area of the cross-section is: • The volume of the approximating cylinder (a disk with thickness ∆x) is: 7.2

  9. Example 2 SOLUTION • The solid lies between x =0 and x = 1. • So, its volume is: 7.2

  10. Example 3 • Find the volume of the solid obtained by rotating the region bounded by y = x3, y =8, and x =0 about the y-axis. • SOLUTION • The region is shown in Figure 7(a) and the resulting solid is shown in Figure 7(b). 7.2

  11. Example 3 SOLUTION • Because 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 • So, the area of a cross-section through y is: • The volume of the approximating cylinder is: 7.2

  12. Example 3 SOLUTION • Since the solid lies between y =0 and y =8, its volume is: 7.2

  13. Example 4 • The region R enclosed by the curves y = x and y = x2 is rotated about the x-axis. Find the volume of the resulting solid. 7.2

  14. Example 4 SOLUTION • The curves y = x and y = x2 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 in Figure 8. 7.2

  15. Example 4 SOLUTION • A cross-section in the plane Pxhas the shape of a washer(an annular ring) with inner radius x2 and outer radius x. 7.2

  16. Example 4 SOLUTION • Thus, we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle: 7.2

  17. Example 4 SOLUTION • Thus, we have: 7.2

  18. Example 5 • Find the volume of the solid obtained by rotating the region in Example 4 about the line y =2. 7.2

  19. Example 5 SOLUTION • The solid and a cross-section are shown in Figure 9. 7.2

  20. Example 5 SOLUTION • Again, the cross-section is a washer. • This time, though, the inner radius is 2 –x and the outer radius is 2 –x2. • The cross-sectional area is: 7.2

  21. Example 5 SOLUTION • So, the volume is: 7.2

  22. Example 6 • Find the volume of the solid obtained by rotating the region in Example 4 about the line x = – 1. 7.2

  23. Example 6 SOLUTION • Figure 11 shows the horizontal cross-section. It is a washer with inner radius 1 + y and outer radius 7.2

  24. Example 6 SOLUTION • So, the cross-sectional area is: 7.2

  25. Example 6 SOLUTION • The volume is: 7.2

  26. Example 7 • Figure 12 shows a solid with a circular base of radius 1. Parallel cross-sections perpendicular to the base are equilateral triangles. Find the volume of the solid. 7.2

  27. Example 7 SOLUTION • Let’s take the circle to be x2+ y2= 1. • The solid, its base, and a typical cross-section at a distance x from the origin are shown in Figure 13. 7.2

  28. Example 7 SOLUTION • As B lies on the circle, we have • So, the base of the triangle ABC is |AB| = 7.2

  29. Example 7 SOLUTION • Since the triangle is equilateral, we see that its height is • Thus, the cross-sectional area is : 7.2

  30. Example 7 SOLUTION • The volume of the solid is: 7.2

  31. Example 8 • Find the volume of a pyramid whose base is a square with side L and whose height is h. • SOLUTION • We place the origin O at the vertex of the pyramid and the x-axis along its central axis as in Figure 14. 7.2

  32. Example 8 SOLUTION • Any plane Px that passes through x and is perpendicular to the x-axis intersects the pyramid in a square with side of length s. 7.2

  33. Example 8 SOLUTION • We can express s in terms of x by observing from the similar triangles in Figure 15 that • Therefore, s = Lx/h • Another method is to observe that the line OPhas slope L/(2h) • So, its equation is y = Lx/(2h) 7.2

  34. Example 8 SOLUTION • Thus, the cross-sectional area is: • The pyramid lies between x =0 and x = h. • So, its volume is: 7.2

More Related