입체도형의 부피

Here is a sketch of the curves $y = \cos(\frac x2)$ and $y = x^2 - \pi^2$.

By inspection, the curves meet at $x = \pm {\pi}$ where both $\cos(\frac x2)$ and $x^2 - \pi^2$ take the value zero. We'll use vertical washers as specified in the question.

• We cut the specified region into thin vertical strips of width $\mathrm{d}x$ as in the figure above.

• When we rotate about the line $y=-\pi^2$, each strip sweeps out a thin washer

• whose inner radius is $r_{in}= (x^2 - {\pi^2} ) - ( {-} {\pi^2} )=x^2$ and outer radius is $r_{out}=\cos(\frac x2) - ( {-} {\pi^2}) =\cos(\frac x2) +\pi^2$, and

• whose thickness is $\mathrm{d}x$ and hence

• whose volume $\pi(r_{out}^2 - r_{in}^2)\mathrm{d}x = \pi\big( {(\cos(\frac x2) +\pi^2)}^2 - {(x^2)}^2\big)\mathrm{d}x$.

• As our leftmost strip is at $x=-\pi$ and our rightmost strip is at $x=\pi$,

the total volume is

$$\begin{aligned} &\pi \int_{-\pi}^{\pi} \left( \cos^2 (\tfrac x2) +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\mathrm{d}x \\ & = \pi \int_{-\pi}^{\pi} \left(\frac{1+\cos(x)}{2} +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\mathrm{d}x \\ \text{Because the integrand is even,} \\ & = 2\pi \int_0^{\pi} \left(\frac{1+\cos(x)}{2} +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\mathrm{d}x \\ & = {2\pi \left[ \frac{1}{2}x + \frac{1}{2}\sin(x) + 4{\pi^2}\sin (\tfrac x2) +{\pi^4} x -\frac{1}{5}x^5 \right]} _0^\pi\\ & = 2\pi \left[\frac{\pi}{2} + 0 + 4{\pi^2} +{\pi^5} - \frac{\pi^5}{5} \right] \\ & = {\pi^2} + 8\pi^3 + \frac{8\pi^6}{5} \end{aligned}$$

We used the fact that the integrand is an even function and the interval of integration $[-\pi, \pi]$ is symmetric, but one can also compute directly.