Cylindrical Coordinates — Question 1

Question 1 – Solution

Strategy. Use $r=\sqrt{x^2+y^2}$ and determine the quadrant before choosing $\theta$.

Step 1: Radius $r=\sqrt{9+27}=6$.

Step 2: Angle Since $x<0$ and $y>0$, the point is in quadrant II. Also $\tan\theta=-\sqrt 3$, so $\theta=2\pi/3$. The height is unchanged. Thus $$\boxed{(r,\theta,z)=(6,2\pi/3,4)}.$$

Step 3: Equivalent forms Adding $2\pi$ gives $(6,8\pi/3,4)$. Reversing the radial sign and adding $\pi$ gives $\boxed{(-6,5\pi/3,4)}$.

Verification. $6\cos(2\pi/3)=-3$, $6\sin(2\pi/3)=3\sqrt 3$, and $z=4$.