Answer
a. $4$
b. $\frac{\pi}{15}$
Work Step by Step
a. Step 1. Draw a diagram as shown in the figure. The limits of integration with respect to $x$ can be found as from $0$ to $4$.
Step 2. The cross section is an isosceles triangle with a height of $h=6$ and a base of $u=\sqrt x-\frac{x}{2}$; we have $dV=\frac{1}{2}uh\ dx=3(\sqrt x-\frac{x}{2})dx$
Step 3. We have
$V=\int_0^4 3(\sqrt x-\frac{x}{2})dx=(2x^{3/2}-\frac{3}{4}x^2)|_0^4=2(4)^{3/2}-\frac{3}{4}(4)^2=2(8)-3(4)=4$
b. In this case, the cross section is a semicircle with radius $r$, where $2r=\sqrt x-\frac{x}{2}$; thus
$dV=\frac{1}{2}\pi r^2 dx=\frac{1}{2}\pi (\frac{\sqrt x-\frac{x}{2}}{2})^2 dx=\frac{\pi}{8} (x-x^{3/2}+\frac{1}{4}x^2) dx$
which leads to
$V=\int_0^4\frac{\pi}{8} (x-x^{3/2}+\frac{1}{4}x^2) dx=\frac{\pi}{8} (\frac{1}{2}x^2-\frac{2}{5}x^{5/2}+\frac{1}{12}x^3)|_0^4=\frac{\pi}{8} (\frac{1}{2}(4)^2-\frac{2}{5}(4)^{5/2}+\frac{1}{12}(4)^3)=\pi (1-\frac{8}{5}+\frac{2}{3})=\frac{\pi}{15}$
