Triple Integrals in Spherical Coordinates — Question 5

Question 5 – Solution

Strategy. The radial and azimuthal factors cancel from the ratio, leaving only the polar-angle factor $1-\cos\alpha$.

Step 1: Sector volume

$$V_{\mathrm{sector}}=\int_0^{2\pi}\int_0^\alpha\int_0^R \rho^2\sin\phi\,d\rho\,d\phi\,d\theta =\frac{2\pi R^3}{3}(1-\cos\alpha).$$ The upper half-ball has volume $2\pi R^3/3$. Hence $$1-\cos\alpha=\frac 13,\qquad \boxed{\alpha=\arccos\!\left(\frac 23\right)}.$$

Step 2: Cartesian cone On $\phi=\alpha$, $z=s\cot\alpha$, where $s=\sqrt{x^2+y^2}$. Since $\sin\alpha=\sqrt 5/3$, we have $\cot\alpha=2/\sqrt 5$. Thus $$\boxed{z=\frac{2}{\sqrt 5}\sqrt{x^2+y^2}}.$$

Verification Substituting $\cos\alpha=2/3$ makes the angular factor $1-\cos\alpha=1/3$, so the sector volume is exactly one-third of the upper half-ball volume.