Question

The base of a solid in the region bounded by the two parabolas y2 = 8x and x2 = 8y. Cross sections of the solid perpendicular to the x-axis are semicircles. What is the volume, in cubic units, of the solid?


288 times pi over 35

576 times pi over 35

144 times pi over 35

8π

Answer 1

The two curves intersect at two points, (0, 0) and (8, 8):

$x^2=8y\implies y=\dfrac{x^2}8$

$y^2=\dfrac{x^4}{64}=8x\implies\dfrac{x^4}{64}-8x=0\implies\dfrac{x(x-8)(x^2+8x+64)}{64}=0$

$\implies x=0,x=8\implies y=0,y=8$

The area of a semicircle with diameter $d$ is $\dfrac{\pi d^2}8$. The diameter of each cross-section is determined by the vertical distance between the two curves for any given value of $x$ between 0 and 8. Over this interval, $y^2=8x\implies y=\sqrt{8x}$ and $\sqrt{8x}>\dfrac{x^2}8$, so the volume of this solid is given by the integral

$\displaystyle\frac\pi8\int_0^8\left(\sqrt{8x}-\dfrac{x^2}8\right)^2\,\mathrm dx=\frac{288\pi}{35}$