Answer
a. $25\pi$
b.$\frac{3}{8\pi}\ unit/sec$
Work Step by Step
a. Draw a diagram as shown in the figure. With the enclosed region revolving around the y-axis, using cylinder approximation, we have
$V=\int_0^{\sqrt {10}} 2\pi x(5-\frac{x^2}{2})dx=\pi \int_0^{\sqrt {10}} (10x-x^3)dx=\pi (5x^2-\frac{1}{4}x^4)|_0^{\sqrt {10}} =\pi (5(\sqrt {10})^2-\frac{1}{4}(\sqrt {10})^4)=25\pi$. Or we can use disk approximation over $dy$ as $V=\int_0^5\pi(x^2)dy=\int_0^5\pi(2y)dy=\pi(y^2)|_0^5=25\pi$
b. Rewrite the second integral above as $V(h)=2\pi\int_0^h(y)dy$ where $h$ is the height of the water level. We have $\frac{dV}{dh}=2\pi h$ and $\frac{dV}{dt}=2\pi h \frac{dh}{dt}$. Given $\frac{dV}{dt}=3$ and at $h=4$, we get $\frac{dh}{dt}=\frac{3}{2\pi(4)}=\frac{3}{8\pi}\ unit/sec$.
