Find the volume of y=sqrt[25-x^2], y=0, x=2, x=4 about the x axis.
Integrating that becomes a mess when I tried it, and it gets a wacky integral that can't be right.
I assume that you want to calculate the volume if the given area rotates around the x-axis.
$\displaystyle V_{rot}=\pi \cdot r^2 \cdot dx$ with r = y.
Then you have:
$\displaystyle V_{rot}=\int_2^4\left(\pi \cdot \left(\sqrt{25-x^2}\right)^2 \right) dx = \pi \left [25x-\dfrac13x^3\right]_2^4 = \dfrac{94}3 \pi$