Exercises: Shell Method

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Exercises for using the shell method.

The region defined by the inequalities $\sqrt {1-x^2} \leq y \leq 1$ for $0 \leq y \leq 1$ is revolved around the $y$ -axis. Compute the volume of the resulting solid using the shell method.

When the slicing variable is $x$ , the radius of a shell is the distance from an $x$ -slice to the axis of rotation. Thus $r(x) = \answer {x} - \answer {0}.$
The height of an $x$ -slice is equal to $h(x) = \answer {1 - \sqrt {1 - x^2}}.$
The volume is equal to the integral of $2 \pi r h$ , so $V = \int _{\answer {0}}^{\answer {1}} \answer {2 \pi x ( 1 - \sqrt {1-x^2}) } dx = \answer {\frac {\pi }{3}}.$ (Note: to compute the integral, split it into two parts and make the substitution $u = 1-x^2$ for one of them.)

The region in the plane bounded above by the graph $y = \sqrt {1+x^2}$ , below by $y = -1 + x + \sqrt {1+x^2}$ , and on the left by $x = 0$ is revolved around the axis $x = 1$ . Compute the volume of the resulting solid using the shell method.

When the slicing variable is $x$ , the radius of a shell is the distance from an $x$ -slice to the axis $x = 0$ . Thus $r(x) = \answer {1} - \answer {x}.$
The height of an $x$ -slice is equal to $h(x) = \answer {-1+x}.$
The volume is equal to the integral of $2 \pi r h$ , so $V = \int _{\answer {-1}}^{\answer {1}} \answer {2 \pi (1-x)^2 } dx = \answer {\frac {16\pi }{3}}.$

The region in the plane $y = \sqrt {x}$ , $y=0$ , and $x = 1$ is revolved around the $y$ -axis. Use the shell method to compute the volume. $V = \answer {\frac {4 \pi }{5}}.$

The same region as above (bounded by $y = \sqrt {x}$ , $y=0$ , and $x = 1$ ) is revolved around the axis $x=1$ . Use the shell method to compute the volume. $V = \answer {\frac {8 \pi }{15}}.$

The same region as above (bounded by $y = \sqrt {x}$ , $y=0$ , and $x = 1$ ) is revolved around the $x$ -axis. Use the shell method to compute the volume.

The “height” of a shell is $1-y^2$ in this case.

$V = \answer {\frac {\pi }{2}}.$

For the same region as above (bounded by $y = \sqrt {x}$ , $y=0$ , and $x = 1$ ), use the shell method to compute the volume when revolved around the axis $y = 1$ . $V = \answer {\frac {5 \pi }{6}}.$

Sample Quiz Questions

The region in the plane bounded below by the curve $y=-x^2$ , above by the curve $y=x^2+2x+2$ , on the right by the line $x = 0$ , and on the left by the line $x = -2$ is revolved around the axis $x = -2$ . Compute the volume of the resulting solid. (Hints won’t reveal until after you choose a response.)

$4\pi$ $6\pi$ $8\pi$ $10\pi$ $12\pi$ $14\pi$

The axis $x = -2$ is parallel to the direction of slices using the integration variable $x$ , which indicates the shell method. The region lies to the right of the axis, which must be the case because the interval $-2 \leq x \leq 0$ lies to the right of the axis $x = -2$ .

The integral to compute equals $\begin {aligned} V &= \int _{-2}^{0}2 \pi (x-(-2))((x^2+2x+2)-(-x^2))~ dx\\ & = \pi \int _{-2}^{0} (4x^3+12x^2+12x+8)~ dx\\ & = \pi \left . \left (x^4+4x^3+6x^2+8x\right ) \right |_{-2}^{0} = 8\pi . \end {aligned}$

The region in the plane bounded on the left by the curve $x=-y^2+4y+1$ , on the right by the curve $x=y^2+2y+1$ , and below by the line $y = -1$ is revolved around the axis $y = -1$ . Compute the volume of the resulting solid. (Hints won’t reveal until after you choose a response.)

$\pi$ $5\pi$ $9\pi$ $13\pi$ $17\pi$ $21\pi$

The axis $y = -1$ is parallel to the direction of slices using the integration variable $y$ , which indicates the shell method. The lower endpoint of integration will be $y = -1$ ; the upper endpoint can be determined by setting $-y^2+4y+1 = y^2+2y+1$ and choosing the solution which is greater than $-1$ . This gives the range $-1 \leq y \leq 0$ . The region lies above the axis, which must be the case because the interval $-1 \leq y \leq 0$ lies above the axis $y = -1$ .

The integral to compute equals $\begin {aligned} V &= \int _{-1}^{0}2 \pi (y-(-1))((y^2+2y+1)-(-y^2+4y+1))~ dy\\ & = \pi \int _{-1}^{0} (4y^3-4y)~ dy\\ & = \pi \left . \left (y^4-2y^2\right ) \right |_{-1}^{0} = 1\pi . \end {aligned}$

The region in the plane between the $x$ -axis and the graph $y = \frac {1}{2 \sqrt {\frac {x^{2}}{3} + 1}}$ in the range $0 \leq x \leq 3$ is revolved around the axis $x = 0$ . Compute the volume of the resulting solid. (Hints won’t reveal until after you choose a response.)

$\displaystyle \frac {3}{2} \pi$ $\displaystyle \frac {8}{5} \pi$ $\displaystyle \frac {5}{3} \pi$ $\displaystyle 2 \pi$ $\displaystyle 3 \pi$ $\displaystyle 5 \pi$

If the variable $x$ is used for slicing, then slices are parallel to the axis of rotation, which indicates the shell method should be used. The radius of a shell is $x$ . The height of a shell is exactly $\frac {1}{2 \sqrt {\frac {x^{2}}{3} + 1}}$ .

The volume of the region is therefore given by $\int _{0}^{3} \frac {\sqrt {3} \pi x}{\sqrt {x^{2} + 3}}\, dx.$

To compute the integral we can use the substitution $u = x^{2} + 3$ which implies the equality $du = \left (2 x\right )dx$ for the differentials. This gives the equality $\begin {aligned} \int \frac {\sqrt {3} \pi x}{\sqrt {x^{2} + 3}}\, dx & = \int \frac {\sqrt {3} \pi }{2 \sqrt {u}}\, du \\ & = \sqrt {3} \pi \sqrt {u}. \end {aligned}$ Reversing the substitution gives $\begin {aligned} \int _{0}^{3} \frac {\sqrt {3} \pi x}{\sqrt {x^{2} + 3}}\, dx & = \left . \left [\sqrt {3} \pi \sqrt {x^{2} + 3} \right ] \right |_{0}^{3}\\ & = \left (6 \pi \right ) - \left (3 \pi \right ) = 3 \pi . \end {aligned}$