Cylindrical Coordinates — Question 1

Section 13-12 · Calculus pages 2601–2602

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Question 1

Convert the rectangular point (−3,33,5)(-3,3\sqrt 3,5) to cylindrical coordinates using the convention r≥0r\ge 0 and 0≤θ<2π0\le\theta<2\pi. Verify by converting back.

Question 1 – Solution

Strategy Compute rr first, then place the angle in the correct quadrant.

Diagram shown in the original SVG worksheet.

Conversion r=(−3)2+(33)2=6.r=\sqrt{(-3)^2+(3\sqrt 3)^2}=6. Because x<0x<0 and y>0y>0, the point lies in Quadrant II. With tan⁡θ=−3\tan\theta=-\sqrt 3, θ=2π/3\theta=2\pi/3. Hence (r,θ,z)=(6,2π3,5).\boxed{(r,\theta,z)=\left(6,\frac{2\pi}{3},5\right)}.

Verification x=6cos⁡(2π/3)=−3x=6\cos(2\pi/3)=-3, y=6sin⁡(2π/3)=33y=6\sin(2\pi/3)=3\sqrt 3, and z=5z=5.