Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curves $y = \cos x$, $x = \pm\pi$, and $y = -1$ about the line $y = -1$.

We shift coordinates so the axis of rotation becomes the baseline, then apply the disk method.

Step 1: Distance to the axis

Since the region lies between $y=\cos x$ (above) and $y=-1$ (below, touching at $x=\pm\pi$), the radius at each $x$ is $r(x)=\cos x-(-1)=\cos x+1\ge0$.

Step 2: Disk method

\[A(x) = \pi(\cos x+1)^2 = \pi(\cos^2x+2\cos x+1)\]

Step 3: Set up the integral

\[V = \pi\int_{-\pi}^{\pi}\cos^2x\,dx + 2\pi\int_{-\pi}^{\pi}\cos x\,dx + \pi\int_{-\pi}^{\pi}1\,dx\]

Step 4: Evaluate each piece

Using $\cos^2x=\frac{1+\cos2x}{2}$: $\displaystyle\int_{-\pi}^{\pi}\cos^2x\,dx = \frac12\left[x+\frac{\sin2x}{2}\right]_{-\pi}^{\pi} = \pi$.

\[\int_{-\pi}^{\pi}\cos x\,dx = [\sin x]_{-\pi}^{\pi} = 0\]

\[\int_{-\pi}^{\pi}1\,dx = 2\pi\]

Step 5: Calculate the volume

\[V = \pi\cdot\pi + 2\pi\cdot0 + \pi\cdot2\pi = \pi^2+2\pi^2 = 3\pi^2\]

\[\boxed{V = 3\pi^2 \text{ cubic units}}\]

Step 6: Numerical value

\[V = 3\pi^2 \approx 29.61 \text{ cubic units}\]

Summary

\[\boxed{V = 3\pi^2 \approx 29.61 \text{ cubic units}}\]

Key insight: Rotating about the cosine curve's own minimum value turns the varying radius into $\cos x+1$, which ranges from $0$ (at $x=\pm\pi$, where the solid pinches to a point) up to $2$ (at $x=0$); the odd $\cos x$ term integrates away by symmetry, leaving a clean $3\pi^2$.