$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\bigmath }[1]{\displaystyle #1} \newcommand {\choicebreak }[0]{} \newcommand {\type }[0]{} \newcommand {\notes }[0]{} \newcommand {\keywords }[0]{} \newcommand {\offline }[0]{} \newcommand {\comments }[0]{\begin {feedback}} \newcommand {\multiplechoice }[0]{\begin {multipleChoice}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

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 horizontalvertical distance from an $x$-slice to the axis of rotation. Thus
• The height of an $x$-slice is equal to
• The volume is equal to the integral of $2 \pi r h$, so (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 horizontalvertical distance from an $x$-slice to the axis $x = 0$. Thus
• The height of an $x$-slice is equal to
• The volume is equal to the integral of $2 \pi r h$, so
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.
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.
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.
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 $y = 1$.

### 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 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 region in the plane between the $x$-axis and the graph 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.)

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