Volume Inside a Sphere and Above a Cone (Spherical Coordinates): 04-BS-1, Undated paper, Q5

Question 5: Volume Inside a Sphere and Above a Cone (Spherical Coordinates) (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.

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The source text calls $x^2+y^2+z^2=9$ an “ellipsoid”, but this equation is a sphere of radius 3 (all three semi-axes equal); the solution below treats it as the sphere it actually is.

Given. The solid region bounded above by the sphere $\rho=3$ and below (in angle) by the cone $z=r$, i.e. the region above the cone and inside the sphere.

Find. The volume of this region.

Approach. Use spherical coordinates: the sphere is $\rho=3$ and the cone $z=\sqrt{x^2+y^2}$ is the constant-$\phi$ surface $\phi=\pi/4$ (since $z=r$ means $\tan\phi=r/z=1$). Integrate $\rho^2\sin\phi$ over $0\le\rho\le3$, $0\le\phi\le\pi/4$, $0\le\theta\le2\pi$.

Solid region inside the sphere $\rho=3$ and above the cone $z=\sqrt{x^2+y^2}$ ($\phi\le\pi/4$).

Identify the cone as a constant-$\phi$ surface. In spherical coordinates $z=\rho\cos\phi$, $r=\rho\sin\phi$; the cone $z=r$ becomes $\cos\phi=\sin\phi\Rightarrow\phi=\pi/4$. “Above the cone” means $0\le\phi\le\pi/4$.
Set up the triple integral. $$V=\int_0^{2\pi}\int_0^{\pi/4}\int_0^3\rho^2\sin\phi\;d\rho\,d\phi\,d\theta.$$
Integrate over $\rho$ and $\theta$. $$\int_0^3\rho^2\,d\rho=9,\qquad\int_0^{2\pi}d\theta=2\pi\ \Rightarrow\ V=18\pi\int_0^{\pi/4}\sin\phi\,d\phi.$$
Integrate over $\phi$ and combine. $$\int_0^{\pi/4}\sin\phi\,d\phi=\big[-\cos\phi\big]_0^{\pi/4}=1-\tfrac{\sqrt2}{2}.$$ $$V=18\pi\left(1-\tfrac{\sqrt2}{2}\right)=\boxed{18\pi-9\sqrt2\,\pi\approx16.56}$$

Quantity	Result
Cone as $\phi=$const	$\phi=\pi/4$
Volume	$18\pi-9\sqrt2\,\pi\approx16.56$

Concept & Theory

Topic. Volumes in spherical coordinates bounded by a sphere and a cone.

Key relations. $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$; a right circular cone with half-angle $\alpha$ measured from the $z$-axis is the surface $\phi=\alpha$; the cone $z=r$ (45° half-angle) is $\phi=\pi/4$.

Why this works. Both bounding surfaces (a sphere centred at the origin and a cone with apex at the origin) are each described by holding exactly one spherical coordinate constant, so the region becomes a coordinate box $0\le\rho\le3,\ 0\le\phi\le\pi/4,\ 0\le\theta\le2\pi$ and the triple integral separates into three single integrals.

Common pitfall. Mislabelling $x^2+y^2+z^2=9$ as an ellipsoid (flagged above) and, separately, integrating $\phi$ from $\pi/4$ to $\pi$ instead of $0$ to $\pi/4$ (which would give the region below the cone).

Source. Stewart, Calculus: Early Transcendentals, Ch. 15 (§15.8 Triple Integrals in Spherical Coordinates).

Question 5 of 7: Volume Inside a Sphere and Above a Cone (Spherical Coordinates)

Question 5: Volume Inside a Sphere and Above a Cone (Spherical Coordinates) (20 marks)