Problem Set: Determining Volumes by Slicing

1. Derive the formula for the volume of a sphere using the slicing method.

2. Use the slicing method to derive the formula for the volume of a cone.

3. Use the slicing method to derive the formula for the volume of a tetrahedron with side length [latex]a.[/latex]

4. Use the disk method to derive the formula for the volume of a trapezoidal cylinder.

5. Explain when you would use the disk method versus the washer method. When are they interchangeable?

For the following exercises (6-10), draw a typical slice and find the volume using the slicing method for the given volume.

6. A pyramid with height 6 units and square base of side 2 units, as pictured here.

This figure is a pyramid with base width of 2 and height of 6 units.

7. A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here.

This figure is a pyramid with base width of 2, length of 3, and height of 4 units.

8. A tetrahedron with a base side of 4 units, as seen here.

This figure is an equilateral triangle with side length of 4 units.

9. A pyramid with height 5 units, and an isosceles triangular base with lengths of 6 units and 8 units, as seen here.

This figure is a pyramid with a triangular base. The view is of the base. The sides of the triangle measure 6 units, 8 units, and 8 units. The height of the pyramid is 5 units.

10. A cone of radius [latex]r[/latex] and height [latex]h[/latex] has a smaller cone of radius [latex]r\text{/}2[/latex] and height [latex]h\text{/}2[/latex] removed from the top, as seen here. The resulting solid is called a frustum.

This figure is a 3-dimensional graph of an upside down cone. The cone is inside of a rectangular prism that represents the xyz coordinate system. the radius of the bottom of the cone is “r” and the radius of the top of the cone is labeled “r/2”.

For the following exercises (11-16), draw an outline of the solid and find the volume using the slicing method.

11. The base is a circle of radius [latex]a.[/latex] The slices perpendicular to the base are squares.

12. The base is a triangle with vertices [latex](0,0),(1,0),[/latex] and [latex](0,1).[/latex] Slices perpendicular to the xy-plane are semicircles.

13. The base is the region under the parabola [latex]y=1-{x}^{2}[/latex] in the first quadrant. Slices perpendicular to the xy-plane are squares.

14. The base is the region under the parabola [latex]y=1-{x}^{2}[/latex] and above the [latex]x\text{-axis}\text{.}[/latex] Slices perpendicular to the [latex]y\text{-axis}[/latex] are squares.

15. The base is the region enclosed by [latex]y={x}^{2}[/latex] and [latex]y=9.[/latex] Slices perpendicular to the [latex]x[/latex]-axis are right isosceles triangles.

16. The base is the area between [latex]y=x[/latex] and [latex]y={x}^{2}.[/latex] Slices perpendicular to the [latex]x[/latex]-axis are semicircles.

For the following exercises (17-24), draw the region bounded by the curves. Then, use the disk method to find the volume when the region is rotated around the [latex]x[/latex]-axis.

17. [latex]x+y=8,x=0,\text{ and }y=0[/latex]

18. [latex]y=2{x}^{2},x=0,x=4,\text{ and }y=0[/latex]

19. [latex]y={e}^{x}+1,x=0,x=1,\text{ and }y=0[/latex]

20. [latex]y={x}^{4},x=0,\text{ and }y=1[/latex]

21. [latex]y=\sqrt{x},x=0,x=4,\text{ and }y=0[/latex]

22. [latex]y= \sin x,y= \cos x,\text{ and }x=0[/latex]

23. [latex]y=\frac{1}{x},x=2,\text{ and }y=3[/latex]

24. [latex]{x}^{2}-{y}^{2}=9\text{ and }x+y=9,y=0\text{ and }x=0[/latex]

For the following exercises (25-32), draw the region bounded by the curves. Then, find the volume when the region is rotated around the [latex]y[/latex]-axis.

25. [latex]y=4-\frac{1}{2}x,x=0,\text{ and }y=0[/latex]

26. [latex]y=2{x}^{3},x=0,x=1,\text{ and }y=0[/latex]

27. [latex]y=3{x}^{2},x=0,\text{ and }y=3[/latex]

28. [latex]y=\sqrt{4-{x}^{2}},y=0,\text{ and }x=0[/latex]

29. [latex]y=\frac{1}{\sqrt{x+1}},x=0,\text{ and }x=3[/latex]

30. [latex]x= \sec (y)\text{ and }y=\frac{\pi }{4},y=0\text{ and }x=0[/latex]

31. [latex]y=\frac{1}{x+1},x=0,\text{ and }x=2[/latex]

32. [latex]y=4-x,y=x,\text{ and }x=0[/latex]

For the following exercises (33-40), draw the region bounded by the curves. Then, find the volume when the region is rotated around the [latex]x[/latex]-axis.

33. [latex]y=x+2,y=x+6,x=0,\text{ and }x=5[/latex]

34. [latex]y={x}^{2}\text{ and }y=x+2[/latex]

35. [latex]{x}^{2}={y}^{3}\text{ and }{x}^{3}={y}^{2}[/latex]

36. [latex]y=4-{x}^{2}\text{ and }y=2-x[/latex]

37. [T] [latex]y= \cos x,y={e}^{\text{−}x},x=0,\text{ and }x=1.2927[/latex]

38. [latex]y=\sqrt{x}\text{ and }y={x}^{2}[/latex]

39. [latex]y= \sin x\text{,}y=5 \sin x,x=0\text{ and }x=\pi [/latex]

40. [latex]y=\sqrt{1+{x}^{2}}\text{ and }y=\sqrt{4-{x}^{2}}[/latex]

For the following exercises (41-45), draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the [latex]y[/latex]-axis.

41. [latex]y=\sqrt{x},x=4,\text{ and }y=0[/latex]

42. [latex]y=x+2,y=2x-1,\text{ and }x=0[/latex]

43. [latex]y=\sqrt[3]{x}\text{ and }y={x}^{3}[/latex]

44. [latex]x={e}^{2y},x={y}^{2},y=0,\text{ and }y=\text{ln}(2)[/latex]

45. [latex]x=\sqrt{9-{y}^{2}},x={e}^{\text{−}y},y=0,\text{ and }y=3[/latex]

46. Yogurt containers can be shaped like frustums. Rotate the line [latex]y=\frac{1}{m}x[/latex] around the [latex]y[/latex]-axis to find the volume between [latex]y=a\text{ and }y=b.[/latex]

This figure has two parts. The first part is a solid cone. The base of the cone is wider than the top. It is shown in a 3-dimensional box. Underneath the cone is an image of a yogurt container with the same shape as the figure.

47. Rotate the ellipse [latex]({x}^{2}\text{/}{a}^{2})+({y}^{2}\text{/}{b}^{2})=1[/latex] around the [latex]x[/latex]-axis to approximate the volume of a football, as seen here.

This figure has an oval that is approximately equal to the image of a football.

48. Rotate the ellipse [latex]({x}^{2}\text{/}{a}^{2})+({y}^{2}\text{/}{b}^{2})=1[/latex] around the [latex]y[/latex]-axis to approximate the volume of a football.

49. A better approximation of the volume of a football is given by the solid that comes from rotating [latex]y= \sin x[/latex] around the [latex]x[/latex]-axis from [latex]x=0[/latex] to [latex]x=\pi .[/latex] What is the volume of this football approximation, as seen here?

This figure has a 3-dimensional oval shape. It is inside of a box parallel to the x-axis on the bottom front edge of the box. The y-axis is vertical to the solid.

50. What is the volume of the Bundt cake that comes from rotating [latex]y= \sin x[/latex] around the [latex]y[/latex]-axis from [latex]x=0[/latex] to [latex]x=\pi ?[/latex]

This figure is a graph of a 3-dimensional solid. It is round, bigger towards the bottom. It has a hole in the center that progressively gets smaller towards the bottom. Next to the graph is an image of a bundt cake, resembling the solid.

For the following exercises (51-56), find the volume of the solid described.

51. The base is the region between [latex]y=x[/latex] and [latex]y={x}^{2}.[/latex] Slices perpendicular to the [latex]x[/latex]-axis are semicircles.

52. The base is the region enclosed by the generic ellipse [latex]({x}^{2}\text{/}{a}^{2})+({y}^{2}\text{/}{b}^{2})=1.[/latex] Slices perpendicular to the [latex]x[/latex]-axis are semicircles.

53. Bore a hole of radius [latex]a[/latex] down the axis of a right cone and through the base of radius [latex]b,[/latex] as seen here.

This figure is an upside down cone. It has a radius of the top as “b”, center at “a”, and height as “b”.

54. Find the volume common to two spheres of radius [latex]r[/latex] with centers that are [latex]2h[/latex] apart, as shown here.

This figure has two circles that intersect. Both circles have radius “r”. There is a line segment from one center to the other. In the middle of the intersection of the circles is point “h”. It is on the line segment.

55. Find the volume of a spherical cap of height [latex]h[/latex] and radius [latex]r[/latex] where [latex]h<r,[/latex] as seen here.

This figure a portion of a sphere. This spherical cap has radius “r” and height “h”.

56. Find the volume of a sphere of radius [latex]R[/latex] with a cap of height [latex]h[/latex] removed from the top, as seen here.

This figure is a sphere with a top portion removed. The radius of the sphere is “R”. The distance from the center to where the top portion is removed is “R-h”.