Question

Find the volume of the solid generated by revolving about the x-axis the region bounded by the upper half of the ellipse(x^2/a^2)+(y^2/b^2)=1and the x-axis, and thus find the volume of a prolate spheroid. Here a and b are positive constants, with a>b.Volume of the solid of revolution:

Answer 1

The upper half of the ellipse has equation

$y=\dfrac ba\sqrt{a^2-x^2}$

with $-a\le x\le a$, so that the volume of the solid (using the disk method) is

$\displaystyle\frac{\pi b}a\int_{-a}^a\left(\sqrt{a^2-x^2}\right)^2-0^2\,\mathrm dx=\frac{2\pi b}a\int_0^a(a^2-x^2)\,\mathrm dx$

$\displaystyle=\frac{2\pi b}a\left(a^2x-\frac{x^3}3\right)\bigg|_0^a$

$\displaystyle=\boxed{\frac{4\pi a^2b}3}$