Surface Area with Polar Coordinates — Question 7

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Question 7

Problem

Set up the surface area obtained by rotating one petal of r=cos⁡2θr=\cos2\theta, −π/4≤θ≤π/4-\pi/4\le\theta\le\pi/4, about the yy-axis.

See the diagram in the original worksheet below.

Original worksheet page 1: question and worked solution for 3-10-007
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Question 7 – Solution

See the diagram in the original worksheet below.

Solution

  1. Compute the polar arc-length element ds=r2+(drdθ)2dθ.ds=\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta.

  2. Express the radius of rotation as a nonnegative distance: use |rsin⁡θ||r\sin\theta| for the xx-axis and |rcos⁡θ||r\cos\theta| for the yy-axis. Then apply S=2π∫ab(radius to the axis)ds,S=2\pi\int_a^b(\text{radius to the axis})\,ds, over an interval that generates the surface exactly once.

  3. Use radius |x|=|rcos⁡θ||x|=|r\cos\theta| and ds=cos⁡22θ+4sin⁡22θdθds=\sqrt{\cos^22\theta+4\sin^22\theta}d\theta.

  4. On this interval rcos⁡θ≥0r\cos\theta\ge0, so the requested setup is S=2π∫−π/4π/4cos⁡2θcos⁡θcos⁡22θ+4sin⁡22θdθ.\boxed{S=2\pi\int_{-\pi/4}^{\pi/4}\cos2\theta\cos\theta \sqrt{\cos^22\theta+4\sin^22\theta}\,d\theta}.

Original worksheet page 2: question and worked solution for 3-10-007

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