Surface Area — Question 2

Question 2

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

Question 2 – Solution

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