Volume Inside an Ellipsoid and Above a Cone: 04-BS-1, May 2013, 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 $z=\sqrt{x^2+y^2}=r$ (cylindrical radius $r$). The solid is bounded below by the cone and above by the ellipsoid's upper surface.

Find. The volume of the region between the two surfaces.

Figure — solid region between the cone z=√(x²+y²) (blue generators) and the ellipsoid cap x²+y²+4z²=5 (red), meeting at the circle r=1, z=1; volume by a cylindrical-coordinate slab integral.

Approach. Work in cylindrical coordinates. Find where the cone and ellipsoid meet to get the radial limit, then integrate the vertical slab thickness (ellipsoid top minus cone) over the disk of that radius.

Find the intersection radius. Substituting $z=r$ into the ellipsoid equation $r^2+4z^2=5$: $$r^2+4r^2=5 \;\Rightarrow\; r^2=1 \;\Rightarrow\; \boxed{r=1,\ z=1}.$$
Express the bounding surfaces in cylindrical coordinates. Cone: $z=r$. Ellipsoid upper surface: $4z^2=5-r^2\Rightarrow z=\tfrac12\sqrt{5-r^2}$.
Set up the volume integral. $$V=\int_0^{2\pi}\int_0^1\left[\tfrac12\sqrt{5-r^2}-r\right] r\,dr\,d\theta = 2\pi\int_0^1\left[\tfrac12 r\sqrt{5-r^2}-r^2\right]dr.$$
Evaluate the two pieces. $\displaystyle\int_0^1 r^2\,dr=\tfrac13$. For $\int_0^1 r\sqrt{5-r^2}\,dr$, substitute $u=5-r^2,\,du=-2r\,dr$: $$\int_0^1 r\sqrt{5-r^2}\,dr = \left[-\tfrac13(5-r^2)^{3/2}\right]_0^1=-\tfrac13(4^{3/2})+\tfrac13(5^{3/2})=\tfrac13(5\sqrt5-8).$$ So $$V=2\pi\left[\tfrac12\cdot\tfrac13(5\sqrt5-8)-\tfrac13\right] = 2\pi\cdot\frac{5\sqrt5-8-2}{6}=\frac{\pi(5\sqrt5-10)}{3}.$$

$$V=\boxed{\dfrac{\pi(5\sqrt5-10)}{3}\approx 1.236}$$

Quantity	Result
Intersection circle	$r=1,\ z=1$
Volume $V$	$\dfrac{\pi(5\sqrt5-10)}{3}\approx 1.236$ (cubic units)

Concept & Theory

Topic. Volume between two axisymmetric surfaces via cylindrical-coordinate integration.

Key relations. Cylindrical coordinates $x=r\cos\theta,\ y=r\sin\theta$; volume between $z=g(r)$ (upper) and $z=h(r)$ (lower) over a disk: $V=\int_0^{2\pi}\int_0^{r_{\max}}[g(r)-h(r)]\,r\,dr\,d\theta$; substitution $u=a^2-r^2$ for $\int r\sqrt{a^2-r^2}\,dr$.

Why this works. Both bounding surfaces are surfaces of revolution about the $z$-axis (the cone and the ellipsoid are both functions of $r=\sqrt{x^2+y^2}$ alone), so cylindrical coordinates decouple the $\theta$-integral (trivially $2\pi$) from a single radial integral of the slab thickness. The radial upper limit $r=1$ is exactly where the two surfaces cross, found by solving the two equations simultaneously.

Common pitfall. Forgetting the extra factor of $r$ from the polar/cylindrical area element $dA=r\,dr\,d\theta$; also taking the wrong (lower) branch of $z=\pm\sqrt{(5-r^2)/4}$ from the ellipsoid — "above the cone" selects the upper hemisphere-like cap.

Source. Stewart, Calculus: Early Transcendentals, Ch. 15 (Multiple Integrals, §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)