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Calculus IIChapter 2: Applications of IntegralsSection 2-2

Surface Area — Question 2

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Question 2

Curve: y=2+cos⁡xy=2+\cos x on 0≤x≤π0\le x\le\pi.
Axis of rotation: y=−1y=-1.
Task: Set up, but do not evaluate, the surface-area integral.

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Original worksheet page 1: question and worked solution for 2-2-002
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Question 2 – Solution

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Step 1: Find the radius. The radius is the vertical distance from the curve to the axis y=−1y=-1: r=(2+cos⁡x)−(−1)=3+cos⁡x.r=(2+\cos x)-(-1)=3+\cos x. Step 2: Find the arc-length factor. y′=−sin⁡x,ds=1+sin⁡2xdx.y'=-\sin x,\qquad ds=\sqrt{1+\sin^2x}\,dx. Step 3: Substitute into S=2π∫rdsS=2\pi\int r\,ds. S=2π∫0π(3+cos⁡x)1+sin⁡2xdx\boxed{S=2\pi\int_0^\pi(3+\cos x)\sqrt{1+\sin^2x}\,dx}

Original worksheet page 2: question and worked solution for 2-2-002

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