Question 6 of 8: Volume Outside a Cylinder, Inside an Ellipsoid

Question 6: Volume Outside a Cylinder, Inside an Ellipsoid (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Cylinder $x^2+y^2=1$ (radius 1, infinite in $z$); ellipsoid $x^2+y^2+4z^2=4$ ($z=\pm\tfrac12\sqrt{4-r^2}$, $r^2=x^2+y^2$, defined for $r\le2$).

Find. The volume inside the ellipsoid but outside the cylinder.

Approach. Work in cylindrical coordinates. At each radius $r$ between the cylinder ($r=1$) and the ellipsoid's equatorial radius ($r=2$), the ellipsoid caps the solid at $z=\pm\tfrac12\sqrt{4-r^2}$; integrate the full height $2z(r)$ times the circumference element $r\,dr\,d\theta$ over that annulus.

1. Set up the volume integral. The ellipsoid meets $z=0$ at $r=2$, so the annulus of interest is $1\le r\le2$, with full $z$-extent $2\cdot\tfrac12\sqrt{4-r^2}=\sqrt{4-r^2}$ at each $r$: $$V=\int_0^{2\pi}\!\!\int_1^2 r\sqrt{4-r^2}\,dr\,d\theta=2\pi\int_1^2 r\sqrt{4-r^2}\,dr.$$
2. Evaluate the radial integral. Substitute $u=4-r^2,\ du=-2r\,dr$: when $r=1,u=3$; when $r=2,u=0$. $$\int_1^2 r\sqrt{4-r^2}\,dr=\frac12\int_0^3\sqrt u\,du=\frac12\cdot\frac23u^{3/2}\Big|_0^3=\frac13\big(3\sqrt3\big)=\sqrt3.$$
3. Assemble the volume. $$V=2\pi\sqrt3.$$

$$V=\boxed{2\sqrt3\,\pi}\approx10.88$$