Triple Integrals in Spherical Coordinates — Question 6

Question 6 – Solution

Strategy. The shifted sphere becomes $\rho\le 2\cos\phi$, so its nonconstant radial bound encodes the displacement from the origin.

Step 1: Bounds Substitution gives $$\rho^2\le 2\rho\cos\phi.$$ For $\rho\ge 0$, the nontrivial radial interval is $0\le\rho\le 2\cos\phi$, which requires $0\le\phi\le\pi/2$. Also $0\le\theta\le 2\pi$.

Step 2: Evaluate $$\begin{align*} V&=\int_0^{2\pi}\int_0^{\pi/2}\int_0^{2\cos\phi} \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\frac{16\pi}{3}\int_0^{\pi/2}\cos^3\phi\sin\phi\,d\phi =\boxed{\frac{4\pi}{3}}. \end{align*}$$

Verification Completing the square gives $x^2+y^2+(z-1)^2\le 1$, a sphere of radius $1$ centered at $(0,0,1)$. Its ordinary volume is $4\pi/3$, confirming the integral.