# Question #e3d0f

Dec 25, 2017

$9 \pi$

#### Explanation:

while you can use shell method, i think it's easier to rewrite y=x as a function of y and use disc method to rotate about the y-axis.

as a function of y, y=x becomes x=y.

using disc method, the volume of the solid is: $\pi {\int}_{a}^{b} {\left(r \left(y\right)\right)}^{2} \mathrm{dy}$

a and b are the upper and lower bounds of integration. in this problem, $b = 3$ and $a = 0$ (because x=y intersects the axis of rotation at $\left(0 , 0\right)$)

r(y) is the distance between the function and the axis of rotation. in this problem, $r \left(y\right) = y - 0 = y$

plugging in: volume=$\pi {\int}_{0}^{3} {y}^{2} \mathrm{dy}$
$= \pi \left(F \left(3\right) - F \left(0\right)\right)$, where $F \left(y\right) = \frac{1}{3} {y}^{3}$ or the antiderivative of ${y}^{2}$.

$= \pi \left(\frac{1}{3} {\left(3\right)}^{3} - \frac{1}{3} {\left(0\right)}^{3}\right)$
$= \pi \left(9 - 0\right) = 9 \pi$