Question

A solid has its base in the xy-plane, bounded by the x axis, the y axis and the function y=3-x⁵. If cross sections are taken perpendicular to the x axis are semicircles whose diameters are in the xy-plane, what is the volume of this solid?

Answer 1

Attached is a plot of the base with one of the cross sections (at $x=0.6$).

The area of any one cross section is given by $\dfrac{\pi r^2}2$, where $r$ is the radius of the circular cross section. In terms of the sections' diameters $d=2r$, the area would be $\dfrac{\pi\left(\frac d2\right)^2}2=\dfrac{\pi d^2}8$.

Each section's diameter is determined by the vertical distance (in the x-y plane) between the curve $y=3-x^5$ and the x-axis ($y=0$), or simply $d=3-x^5$. So the area of any one cross-section for a given $x$ is $\dfrac{\pi(3-x^5)^2}8$.

The region extends from $x=0$ to $x=3^{1/5}$ (the positive root of $3-x^5$), so the volume of the solid would be

$\displaystyle\int_0^{3^{1/5}}\frac{\pi(3-x^5)^2}8\,\mathrm dx=\frac\pi8\int_0^{3^{1/5}}(3-x^5)^2\,\mathrm dx$

You can compute this by expanding the integrand, then integrating term by term. You should find a volume of $\dfrac{75\times3^{1/5}\pi}{88}\approx3.335$.