Volume Inside a Sphere and Above a Cone (Spherical Coordinates): 04-BS-1, December 2017, Q4

Question 4: 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.

Given. Sphere of radius $2$ centred at the origin; cone $\sqrt3z=r$ (with $r=\sqrt{x^2+y^2}$), i.e. $z=r/\sqrt3$.

Find. The volume of the region satisfying both $x^2+y^2+z^2\leq4$ and $z\geq r/\sqrt3$ (above/inside the cone).

Axial (r–z) cross-section: shaded wedge is the region inside the sphere $r_{\text{sph}}\le2$ and above the cone ($\varphi\le60^\circ$ from the $z$-axis); revolving it about the $z$-axis gives the full solid.

Approach. Convert to spherical coordinates $(\rho,\varphi,\theta)$ with $z=\rho\cos\varphi$, $r=\rho\sin\varphi$. The cone $\sqrt3z=r$ becomes $\sqrt3\cos\varphi=\sin\varphi\Rightarrow\tan\varphi=\sqrt3\Rightarrow\varphi=\pi/3$, and "above the cone" means the smaller cone angle $0\leq\varphi\leq\pi/3$ (closer to the $+z$ axis). Integrate $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$ over $0\leq\rho\leq2$, $0\leq\varphi\leq\pi/3$, $0\leq\theta\leq2\pi$.

Convert the cone to spherical coordinates. $\sqrt3z=r\Rightarrow\sqrt3\rho\cos\varphi=\rho\sin\varphi\Rightarrow\tan\varphi=\sqrt3\Rightarrow\varphi=\dfrac\pi3\ (60^\circ)$. Points "above the cone" (closer to the $z$-axis, larger $z$ for given $r$) satisfy $\varphi\in[0,\pi/3]$.
Set up the triple integral. $$V=\int_0^{2\pi}\int_0^{\pi/3}\int_0^2\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta.$$
Integrate over $\rho$, then $\varphi$, then $\theta$. Since the three integrals separate, $$V=\left(\int_0^2\rho^2d\rho\right)\left(\int_0^{\pi/3}\sin\varphi\,d\varphi\right)\left(\int_0^{2\pi}d\theta\right)=\left[\frac{\rho^3}3\right]_0^2\cdot\bigl[-\cos\varphi\bigr]_0^{\pi/3}\cdot2\pi.$$ $$=\frac83\cdot\left(1-\frac12\right)\cdot2\pi=\frac83\cdot\frac12\cdot2\pi=\frac{8\pi}3.$$

$$V=\boxed{\dfrac{8\pi}3}\approx8.378$$

Quantity	Result
Cone half-angle	$\varphi=60^\circ=\pi/3$
$\rho$-integral	$8/3$
$\varphi$-integral	$1/2$
Volume $V$	$8\pi/3\approx8.378$

Concept & Theory

Topic. Triple integration in spherical coordinates for a sphere/cone region.

Key relations. $x=\rho\sin\varphi\cos\theta,\ y=\rho\sin\varphi\sin\theta,\ z=\rho\cos\varphi$, with volume element $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$. A sphere of radius $R$ is $\rho\le R$; a cone with half-angle $\alpha$ from the $z$-axis is $\varphi=\alpha$ (or $\varphi\le\alpha$ for the interior).

Why this works. Both bounding surfaces are naturally described by constant values of spherical coordinates ($\rho=2$ for the sphere, $\varphi=\pi/3$ for the cone), so the region of integration becomes a simple box $[0,2]\times[0,\pi/3]\times[0,2\pi]$ in $(\rho,\varphi,\theta)$-space, and because the integrand $\rho^2\sin\varphi$ separates into a product of single-variable factors, the triple integral factors into three ordinary integrals.

Common pitfall. Solving $\tan\varphi=\sqrt3$ but picking the wrong root ($\varphi=2\pi/3$ instead of $\pi/3$) or integrating "above the cone" as the region $\varphi\ge\pi/3$ (the wider, outer cone) rather than the narrower cone containing the $z$-axis.

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

Question 4 of 8: Volume Inside a Sphere and Above a Cone (Spherical Coordinates)

Question 4: Volume Inside a Sphere and Above a Cone (Spherical Coordinates) (20 marks)