Spherical Coordinates — Question 10

Question 10 – Solution

Strategy. A nonnegative radial variable can lie below $6\cos\phi$ only when that upper bound is nonnegative.

Step 1: Consistency Since $\rho\ge 0$, we require $6\cos\phi\ge 0$. On $0\le\phi\le\pi$, this means $0\le\phi\le\pi/2$. Intersecting with the proposed range gives $$\boxed{\pi/6\le\phi\le\pi/2}.$$ Angles $\pi/2<\phi\le 2\pi/3$ contribute no points because their proposed upper radial bound is negative.

Step 2: Correct description $$\boxed{0\le\theta<2\pi,\qquad \pi/6\le\phi\le\pi/2,\qquad 0\le\rho\le 6\cos\phi}.$$ The radial surface is the sphere $x^2+y^2+(z-3)^2=9$. The lower angular cutoff removes the portion closer than $\pi/6$ to the positive $z$-axis, leaving the part of the ball outside that narrow cone.

Step 3: Boundary circle On $\phi=\pi/6$ and the sphere, $\rho=3\sqrt 3$. Hence $$z=\rho\cos\phi=\frac 92,\qquad r=\rho\sin\phi=\frac{3\sqrt 3}{2}.$$ The surfaces meet in $$\boxed{x^2+y^2=27/4,\qquad z=9/2}.$$