Volume Inside an Ellipsoid and Above a Cone: 04-BS-1, December 2015, Q8

Question 8: 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 $z=\sqrt{x^2+y^2}=r$ (in cylindrical coordinates), $z\ge0$.

Find. The volume of the region between the cone and the upper ellipsoid surface.

Cross-section: cone $z=r$ meets the ellipsoid $r^2+4z^2=5$ at $r=1,z=1$; the solid fills $r\le1$, from the cone up to the ellipsoid's upper surface.

Approach. Work in cylindrical coordinates. Find the radius $r$ where the cone meets the ellipsoid's upper surface (their common boundary circle), then integrate the vertical extent from the cone up to the ellipsoid over $0\le r\le$ that radius.

Find where the cone meets the ellipsoid. The ellipsoid's upper surface is $z=\sqrt{(5-r^2)/4}$. Setting this equal to the cone $z=r$: $$r^2=\frac{5-r^2}{4} \;\Rightarrow\; 4r^2=5-r^2 \;\Rightarrow\; 5r^2=5 \;\Rightarrow\; r=1\ (z=1).$$ For $0\le r\le1$ the cone lies below the ellipsoid's top surface, so the described solid (inside the ellipsoid, above the cone) is the region swept over that range.
Set up the volume integral. $$V=\int_0^{2\pi}\int_0^1 r\left[\sqrt{\frac{5-r^2}{4}}-r\right]dr\,d\theta = 2\pi\int_0^1\left[\frac r2\sqrt{5-r^2}-r^2\right]dr.$$
Evaluate the two pieces. Substituting $u=5-r^2$, $du=-2r\,dr$: $$\int_0^1 r\sqrt{5-r^2}\,dr=\left[-\frac13(5-r^2)^{3/2}\right]_0^1=-\frac13\big(4^{3/2}-5^{3/2}\big)=\frac{5\sqrt5-8}{3}.$$ So $\displaystyle\int_0^1\frac r2\sqrt{5-r^2}\,dr=\frac{5\sqrt5-8}{6}$, and $\displaystyle\int_0^1 r^2\,dr=\frac13$.
Combine. $$V=2\pi\left[\frac{5\sqrt5-8}{6}-\frac13\right]=2\pi\cdot\frac{5\sqrt5-10}{6}=\frac{\pi(5\sqrt5-10)}{3}.$$

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

Quantity	Result
Cone–ellipsoid intersection radius	$r=1$ (at $z=1$)
$\int_0^1 r\sqrt{5-r^2}\,dr$	$(5\sqrt5-8)/3$
Volume $V$	$\dfrac{5\pi(\sqrt5-2)}{3}\approx1.236$

Concept & Theory

Topic. Volume between two axisymmetric surfaces (a cone and an ellipsoid of revolution) via cylindrical-coordinate integration.

Key relations. For a solid between a lower surface $z=g(r)$ and upper surface $z=h(r)$ (both axisymmetric), $\displaystyle V=2\pi\int_0^{r_0} r\big[h(r)-g(r)\big]\,dr$, where $r_0$ is the radius at which the two surfaces meet.

Why this works. Both bounding surfaces — the cone $z=r$ and the ellipsoid's cap $z=\sqrt{(5-r^2)/4}$ — depend only on $r$, so the solid is a surface of revolution and cylindrical coordinates collapse the problem to a single radial integral times $2\pi$ from the $\theta$-integration. The upper limit $r_0=1$ is found by solving for where the two surfaces coincide (the circle where the cone pinches the ellipsoid cap shut).

Common pitfall. Forgetting the factor of $r$ in $dV=r\,dz\,dr\,d\theta$; also using the wrong branch of the ellipsoid (the lower cap $z=-\sqrt{(5-r^2)/4}$) or integrating $r$ beyond $r_0=1$, where the cone would rise above the ellipsoid and the integrand would go negative.

Source. Stewart, Calculus: Early Transcendentals, Ch. 15 (§15.7 Triple Integrals in Cylindrical Coordinates).

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

Question 8: Volume Inside an Ellipsoid and Above a Cone (20 marks)