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Mathematics LibreTexts

2.2E: Exercises for Section 6.2

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

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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 a.

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?

Volumes by Slicing

For 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.

Solution:
Here the cross-sections are squares taken perpendicular to the y-axis.
We use the vertical cross-section of the pyramid through its center to obtain an equation relating x and y.
Here this would be the equation, y=66x. Since we need the dimensions of the square at each y-level, we solve this equation for x to get, x=1y6.
This is half the distance across the square cross-section at the y-level, so the side length of the square cross-section is, s=2(1y6).
Thus, we have the area of a cross-section is,

A(y)=[2(1y6)]2=4(1y6)2.

Then,V=604(1y6)2dy=2401u2du,whereu=1y6,sodu=16dy,6du=dy=2410u2du=24u33|10=8u3|10=8(1303)=8units3

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.

Answer
V=3232=1623 units3

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 r and height h has a smaller cone of radius r/2 and height h/2 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”.

Answer
V=7π12hr2 units3

For 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 a. The slices perpendicular to the base are squares.

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

Answer

This figure shows the x-axis and the y-axis with a line starting on the x-axis at (1,0) and ending on the y-axis at (0,1). Perpendicular to the xy-plane are 4 shaded semi-circles with their diameters beginning on the x-axis and ending on the line, decreasing in size away from the origin.

V=10π(1x)28dx=π24 units3

13) The base is the region under the parabola y=1x2 in the first quadrant. Slices perpendicular to the xy-plane are squares.

14) The base is the region under the parabola y=1x2 and above the x-axis. Slices perpendicular to the y-axis are squares.

Answer

This figure shows the x-axis and the y-axis in 3-dimensional perspective. On the graph above the x-axis is a parabola, which has its vertex at y=1 and x-intercepts at (-1,0) and (1,0). There are 3 square shaded regions perpendicular to the x y plane, which touch the parabola on either side, decreasing in size away from the origin.

V=104(1y)dy=2 units3

15) The base is the region enclosed by y=x2 and y=9. Slices perpendicular to the x-axis are right isosceles triangles.

16) The base is the area between y=x and y=x2. Slices perpendicular to the x-axis are semicircles.

Answer

This figure is a graph with the x and y axes diagonal to show 3-dimensional perspective. On the first quadrant of the graph are the curves y=x, a line, and y=x^2, a parabola. They intersect at the origin and at (1,1). Several semicircular-shaped shaded regions are perpendicular to the x y plane, which go from the parabola to the line and perpendicular to the line.

V=10π8(xx2)2dx=π240 units3

Disk and Washer Method

For exercises 17 - 24, draw the region bounded by the curves. Then, use the disk or washer method to find the volume when the region is rotated around the x-axis.

17) x+y=8,x=0, and y=0

18) y=2x2,x=0,x=4, and y=0

Answer

This figure is a graph in the first quadrant. It is a shaded region bounded above by the curve y=2x^2, below by the x-axis, and to the right by the vertical line x=4.

V=404πx4dx=4096π5 units3

19) y=ex+1,x=0,x=1, and y=0

20) y=x4,x=0, and y=1

Answer

This figure is a graph in the first quadrant. It is a shaded region bounded above by the line y=1, below by the curve y=x^4, and to the left by the y-axis.

V=10π(12(x4)2)dx=10π(1x8)dx=8π9 units3

21) y=x,x=0,x=4, and y=0

22) y=sinx,y=cosx, and x=0

Answer

This figure is a shaded region bounded above by the curve y=cos(x), below to the left by the y-axis and below to the right by y=sin(x). The shaded region is in the first quadrant.

V=π/40π(cos2xsin2x)dx=π/40πcos2xdx=π2 units3

23) y=1x,x=2, and y=3

24) x2y2=9 and x+y=9,y=0 and x=0

Answer

This figure is a graph in the first quadrant. It is a shaded region bounded above by the line x + y=9, below by the x-axis, to the left by the y-axis, and to the left by the curve x^2-y^2=9.

V=207π units3

For exercises 25 - 32, draw the region bounded by the curves. Then, find the volume when the region is rotated around the y-axis.

25) y=412x,x=0, and y=0

26) y=2x3,x=0,x=1, and y=0

Answer

This figure is a graph in the first quadrant. It is a shaded region bounded above by the curve y=2x^3, below by the x-axis, and to the right by the line x=1.

V=4π5 units3

27) y=3x2,x=0, and y=3

28) y=4x2,y=0, and x=0

Answer

This figure is a graph in the first quadrant. It is a quarter of a circle with center at the origin and radius of 2. It is shaded on the inside.

V=16π3 units3

29) y=1x+1,x=0, and x=3

30) x=sec(y) and y=π4,y=0 and x=0

Answer

This figure is a graph in the first quadrant. It is a shaded region bounded above by the line y=pi/4, to the right by the curve x=sec(y), below by the x-axis, and to the left by the y-axis.

V=π units3

31) y=1x+1,x=0, and x=2

32) y=4x,y=x, and x=0

Answer

This figure is a graph in the first quadrant. It is a shaded triangle bounded above by the line y=4-x, below by the line y=x, and to the left by the y-axis.

V=16π3 units3

For exercises 33 - 40, draw the region bounded by the curves. Then, find the volume when the region is rotated around the x-axis.

33) y=x+2,y=x+6,x=0, and x=5

34) y=x2 and y=x+2

Answer

This figure is a graph above the x-axis. It is a shaded region bounded above by the line y=x+2, and below by the parabola y=x^2.

V=72π5 units3

35) x2=y3 and x3=y2

36) y=4x2 and y=2x

Answer

This figure is a shaded region bounded above by the curve y=4-x^2 and below by the line y=2-x.

V=108π5 units3

37) [T] y=cosx,y=ex,x=0, and x=1.2927

38) y=x and y=x2

Answer

This figure is a graph in the first quadrant. It is a shaded region bounded above by the curve y=squareroot(x), below by the curve y=x^2.

V=3π10 units3

39) y=sinx,y=5sinx,x=0 and x=π

40) y=1+x2 and y=4x2

Answer

This figure is a shaded region bounded above by the curve y=squareroot(4-x^2) and, below by the curve y=squareroot(1+x^2).

V=26π units3

For 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 y-axis.

41) y=x,x=4, and y=0

42) y=x+2,y=2x1, and x=0

Answer

This figure is a graph in the first quadrant. It is a shaded region bounded above by the line y=x+2, below by the line y=2x-1, and to the left by the y-axis.

V=9π units3

43) y=3x and y=x3

44) x=e2y,x=y2,y=0, and y=ln(2)

Answer

This figure is a graph in the first quadrant. It is a shaded region bounded above by the curve y=ln(2), below by the x-axis, to the left by the curve x=y^2, and to the right by the curve x=e^(2y).

V=π20(754ln5(2)) units3

45) x=9y2,x=ey,y=0, and y=3

46) Yogurt containers can be shaped like frustums. Rotate the line y=(1m)x around the y-axis to find the volume between y=a and y=b.

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.

Answer
V=m2π3(b3a3) units3

47) Rotate the ellipse x2a2+y2b2=1 around the x-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 x2a2+y2b2=1 around the y-axis to approximate the volume of a football.

Answer
V=4a2bπ3 units3

49) A better approximation of the volume of a football is given by the solid that comes from rotating y=sinx around the x-axis from x=0 to x=π. 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.

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

51) The base is the region between y=x and y=x2. Slices perpendicular to the x-axis are semicircles.

52) The base is the region enclosed by the generic ellipse x2a2+y2b2=1. Slices perpendicular to the x-axis are semicircles.

Answer
V=2ab2π3 units3

53) Bore a hole of radius a down the axis of a right cone and through the base of radius b, 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 r with centers that are 2h 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.

Answer
V=π12(r+h)2(6rh) units3

55) Find the volume of a spherical cap of height h and radius r where h<r, 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 R with a cap of height h 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”.

Answer
V=π3(h+R)(h2R)2 units3

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


This page titled 2.2E: Exercises for Section 6.2 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.

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