Spherical Coordinates — Question 1

Question 1 – Solution

Strategy. Find radial distance first, use the quadrant of $(x,y)$ for $\theta$, and use $\cos\phi=z/\rho$.

Step 1: Radius $$\rho=\sqrt{x^2+y^2+z^2}=\sqrt{9+9+18}=6.$$

Step 2: Angles The projection $(-3,3)$ lies in quadrant II, so $\theta=3\pi/4$. Also $$\cos\phi=\frac{-3\sqrt 2}{6}=-\frac{\sqrt 2}{2},\qquad 0\le\phi\le\pi,$$ giving $\phi=3\pi/4$. Hence $$\boxed{(\rho,\theta,\phi)=(6,3\pi/4,3\pi/4)}.$$

Step 3: Equivalence and verification Adding $2\pi$ to the azimuth gives $(6,11\pi/4,3\pi/4)$. Moreover, $$6\sin(3\pi/4)\cos(3\pi/4)=-3,\quad 6\sin(3\pi/4)\sin(3\pi/4)=3,\quad 6\cos(3\pi/4)=-3\sqrt 2.$$