Example 15.2.1 Find the volume under $z=\sqrt{4-r^2}$ above the quarter circle bounded by the two axes and the circle $x^2+y^2=4$ in the first quadrant.

In terms of $r$ and $\theta$, this region is described by the restrictions $0\le r\le 2$ and $0\le\theta\le\pi/2$, so we have $$\eqalign{ \int_{0}^{\pi/2}\int_{0}^{2} \sqrt{4-r^2}\;r\,dr\,d\theta &=\int_{0}^{\pi/2}\left. -{1\over3}(4-r^2)^{3/2}\right|_0^2\,d\theta\cr &=\int_{0}^{\pi/2} {8\over3}\,d\theta\cr &={4\pi\over3}.\cr }$$ The surface is a portion of the sphere of radius 2 centered at the origin, in fact exactly one-eighth of the sphere. We know the formula for volume of a sphere is $(4/3)\pi r^3$, so the volume we have computed is $(1/8)(4/3)\pi 2^3=(4/3)\pi$, in agreement with our answer. (From another point of view, what we've done is prove that the volume of a sphere of radius 2 is $(32/3)$. If you replace 2 by $a$ and do the integral again, it is not any more difficult, and you will prove that the volume of a sphere of radius $a$ is $(4/3)\pi a^3$.) $\square$

Example 15.2.2 Find the volume under $z=\sqrt{4-r^2}$ above the region enclosed by the curve $r=2\cos\theta$, $-\pi/2\le\theta\le\pi/2$; see figure 15.2.2. The region is described in polar coordinates by the inequalities $-\pi/2\le\theta\le\pi/2$ and $0\le r\le2\cos\theta$, so the double integral is $$\int_{-\pi/2}^{\pi/2}\int_{0}^{2\cos\theta} \sqrt{4-r^2}\;r\,dr\,d\theta =2\int_{0}^{\pi/2}\int_{0}^{2\cos\theta} \sqrt{4-r^2}\;r\,dr\,d\theta.$$ We can rewrite the integral as shown because of the symmetry of the volume; this avoids a complication during the evaluation. Proceeding: $$\eqalign{ 2\int_{0}^{\pi/2}\int_{0}^{2\cos\theta} \sqrt{4-r^2}\;r\,dr\,d\theta &=2\int_{0}^{\pi/2}-{1\over3}\left.(4-r^2)^{3/2}\right|_0^{2\cos\theta}\,d\theta\cr &=2\int_{0}^{\pi/2}-{8\over3}\sin^3\theta+{8\over3}\,d\theta\cr &=\left.2\left(-{8\over3}{\cos^3\theta\over3}-\cos\theta+{8\over3}\theta\right)\right|_0^{\pi/2}\cr &={8\over3}\pi-{32\over9}.\cr }$$ $\square$