Question 6 of 8: Volume Inside an Ellipsoid and Above a Cone

Question 6: Volume Inside an Ellipsoid and Above a Cone (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. Ellipsoid $x^2+y^2+4z^2=5$; cone $z=\sqrt{x^2+y^2}=r$ (cylindrical coordinates).

Find. The volume of the region inside the ellipsoid and above the cone.

Approach. Work in cylindrical coordinates. Find the crossing radius where the cone meets the ellipsoid's upper cap, then integrate the vertical gap (ellipsoid cap minus cone) over the disk of that radius.

1. Crossing radius. Set the cone equal to the ellipsoid's cap: $r=\sqrt{(5-r^2)/4}\Rightarrow4r^2=5-r^2\Rightarrow5r^2=5\Rightarrow r=1$ (for $r\ge0$).
2. Set up the volume integral. For $0\le r\le1$, $z$ runs from the cone $z=r$ up to the ellipsoid cap $z=\sqrt{(5-r^2)/4}$: $$V=\int_0^{2\pi}\!\!\int_0^1\left[\sqrt{\tfrac{5-r^2}4}-r\right]r\,dr\,d\theta=2\pi\int_0^1\left[\tfrac r2\sqrt{5-r^2}-r^2\right]dr.$$
3. Integrate the ellipsoid term. With $u=5-r^2$, $du=-2r\,dr$: $$\int_0^1\tfrac r2\sqrt{5-r^2}\,dr=-\tfrac16\left[u^{3/2}\right]_{u=5}^{u=4}=-\tfrac16\left(4^{3/2}-5^{3/2}\right)=\tfrac{5\sqrt5-8}6.$$
4. Integrate the cone term and combine. $\int_0^1r^2\,dr=\tfrac13$. So $$\int_0^1\left[\tfrac r2\sqrt{5-r^2}-r^2\right]dr=\frac{5\sqrt5-8}6-\frac13=\frac{5\sqrt5-10}6.$$ $$V=2\pi\cdot\frac{5\sqrt5-10}6=\frac{5\pi(\sqrt5-2)}3.$$

$$V=\boxed{\dfrac{5\pi(\sqrt5-2)}3}\approx1.236$$