The base of a solid in the region bounded by the two parabolas $y^2 = 8x$ and $x^2 = 8y$. Cross sections of the solid perpendicular to the x-axis are semicircles. What is the volume, in cubic units, of the solid?
The answer choices are:
A) $\frac{288\pi}{35}$ B) $\frac{576\pi}{35}$ C) $\frac{144\pi}{35}$ D) $8\pi$
I started off by writing the integral like this:
$$\int_{0}^{8} \frac{1}{2}\pi\bigg(\frac{1}{2}\bigg(\sqrt{8x}-\frac{x^2}{8}\bigg) \bigg)^2\, dx$$
then I simplified it to
$$\frac{\pi}{8}\int_{0}^{8} \bigg(\sqrt{8x}-\frac{x^2}{8}\bigg)^2\, dx$$
I can't think of anyway to solve this problem from here, without using a calculator.
Expand the binomial term: $$V = \frac{\pi}{8} \int_{x=0}^8 \left(\sqrt{8x} - \frac{x^2}{8}\right)^2 \, dx = \frac{\pi}{8} \int_{x=0}^8 8x - 2\sqrt{8x} \cdot \frac{x^2}{8} + \frac{x^4}{64} \, dx.$$ Now integrate term by term: $$V = \frac{\pi}{8} \left[4x^2 - \frac{\sqrt{2}}{7} x^{7/2} + \frac{x^5}{5(64)} \right]_{x=0}^8.$$