입체도형의 부피

(a) The top and the bottom of the circle have equations $y=T(x)=2+\sqrt{1-x^2}$ and $y=B(x) = 2-\sqrt{1-x^2}$, respectively.

enterline

When $R$ is rotated about the $x$–axis, the vertical strip of $R$ in the figure above sweeps out a washer with thickness $\mathrm{d}x$, outer radius $T(x)$ and inner radius $B(x)$. This washer has volume

$$\pi\big(T(x)^2- B(x)^2\big)\,\mathrm{d}x = \pi\big(T(x)+ B(x)\big)\big(T(x)- B(x)\big)\,\mathrm{d}x = \pi\times 4\times 2\sqrt{1-x^2}\,\mathrm{d}x$$

Hence the volume of the solid is

$$8\pi\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x$$

(b) Since $y=\sqrt{1-x^2}$ is equivalent to $x^2+y^2=1$, $y\ge 0$, the integral is $8\pi$ times the area of the upper half of the circle $x^2+y^2=1$ and hence is $8\pi\times \frac{1}{2}\pi 1^2 = 4\pi^2$.