
Exercises for the disk and washer methods.

The region $0 \leq y \leq \sqrt {x}$ with $x \leq 1$, shown below, is revolved around the $x$-axis. Use the disk method to find the volume of the solid of revolution.

The radius $R(x)$ will be a difference of $y$-values because slices are indexed by the variable $x$. Each slice will extend from $y=0$ to $y = \sqrt {x}$, and so $R(x)$ must be the larger of these $y$-values minus the smaller of these $y$-values.
The region $0 \leq y \leq \sqrt {x}$ with $x \leq 1$, shown below, is revolved around the axis $x=1$. Use the disk method to find the volume of the solid of revolution.

The radius $R(y)$ will be a difference of $x$-values because slices are indexed by the variable $y$. Each slice will extend from $x = y^2$ to $x = 1$, and so $R(y)$ must be the larger of these $x$-values minus the smaller of these $x$-values
The region $0 \leq y \leq \sqrt {x}$ with $x \leq 1$, shown below, is revolved around the axis $x=0$. Use the washer method to find the volume of the solid of revolution.

Each radius will be a difference of $x$-values because slices are indexed by the variable $y$. The distance from the axis $x=0$ to the line $x=1$ is $1$, and the distance from the axis $x=0$ to $x = y^2$ is $y^2$.
The region $0 \leq y \leq \sqrt {x}$ with $x \leq 1$, shown below, is revolved around the axis $y=1$. Use the washer method to find the volume of the solid of revolution.

Each radius will be a difference of $y$-values because slices are indexed by the variable $x$. The distance from the axis $y=1$ to the line $y=0$ is $1$, and the distance from the axis $y=1$ to $y = \sqrt {x}$ is $1 - \sqrt {x}$.

### Sample Quiz Questions

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

$16\pi$ $20\pi$ $24\pi$ $28\pi$ $32\pi$ $36\pi$

The region in the plane bounded below by the curve $y=-2x^2+5x+2$, above by the curve $y=-2x^2+2x+2$, on the right by the line $x = 0$, and on the left by the line $x = -1$ is revolved around the axis $y = 2$. Compute the volume of the resulting solid. (Hints won’t reveal until after you choose a response.)

$10\pi$ $14\pi$ $18\pi$ $22\pi$ $26\pi$ $30\pi$

The region in the plane given by $\left |{- \frac {x}{2} + \frac {1}{2} \sqrt {9 - 6 x^{2}}}\right | \leq y \leq \frac {x}{2} + \frac {1}{2} \sqrt {9 - 6 x^{2}}$ and $0 \leq x \leq \frac {2}{3} \sqrt {3}$ is revolved around the $x$-axis. Compute the volume of the resulting solid. (Hints won’t reveal until after you choose a response.)

$\displaystyle \frac {13}{9} \pi$ $\displaystyle \frac {19}{9} \pi$ $\displaystyle \frac {26}{9} \pi$ $\displaystyle \frac {28}{9} \pi$ $\displaystyle \frac {37}{9} \pi$ $\displaystyle \frac {49}{9} \pi$