By converting to polar coordinates, calculate

By converting to polar coordinates, calculate the following integral over each of the two regions described below: $$ \iint_D \sqrt{a^2-x^2-y^2}\,dx\,dy. $$ $1^{\circ}$ $D$ is bounded by the circle $x^2+y^2=a^2$ and the lines $y=x$ and $y=x\sqrt3$.
$2^{\circ}$ $D$ is the disk $x^2+y^2-ax\leqslant0$.

Part $1^\circ$ · Sector between two lines

Step 1 · Identify the region

The region is the sector of the disk $x^2+y^2\leqslant a^2$ between the line $y=x$ (angle $\theta=\frac\pi4$) and the line $y=x\sqrt3$ (angle $\theta=\arctan\sqrt3=\frac\pi3$): that is, $\frac\pi4\leqslant\theta\leqslant\frac\pi3$, $0\leqslant r\leqslant a$.

Step 2 · Set up and integrate with respect to $r$

$$ I_1 = \int_{\pi/4}^{\pi/3}\int_0^a \sqrt{a^2-r^2}\,r\,dr\,d\theta. $$

With $u=a^2-r^2$, $du=-2r\,dr$:

$$ \int_0^a r\sqrt{a^2-r^2}\,dr = \frac12\int_0^{a^2}\sqrt u\,du = \frac12\left[\frac{2u^{3/2}}{3}\right]_0^{a^2} = \frac{a^3}{3}. $$

Step 3 · Integrate with respect to $\theta$

$$ I_1 = \int_{\pi/4}^{\pi/3}\frac{a^3}{3}\,d\theta = \frac{a^3}{3}\left(\frac\pi3-\frac\pi4\right) = \frac{a^3}{3}\cdot\frac{\pi}{12} = \frac{\pi a^3}{36}. $$

Part $2^\circ$ · Inside the circle $x^2+y^2=ax$

Step 1 · Convert the circle to polar form

$x^2+y^2-ax=0$ completes the square to $\left(x-\frac a2\right)^2+y^2=\frac{a^2}{4}$, a circle of radius $\frac a2$ centered at $\left(\frac a2,0\right)$. In polar form: $r^2=ar\cos\theta \Rightarrow r=a\cos\theta$, traced for $-\frac\pi2\leqslant\theta\leqslant\frac\pi2$ (where $\cos\theta\geqslant0$).

Step 2 · Integrate with respect to $r$

$$ I_2 = \int_{-\pi/2}^{\pi/2}\int_0^{a\cos\theta} \sqrt{a^2-r^2}\,r\,dr\,d\theta. $$

As in Part $1^\circ$, with $R=a\cos\theta$:

$$ \int_0^{R} r\sqrt{a^2-r^2}\,dr = \frac13\Big[a^3-(a^2-R^2)^{3/2}\Big]. $$

Here $a^2-R^2 = a^2-a^2\cos^2\theta = a^2\sin^2\theta$, so:

$$ (a^2-R^2)^{3/2} = \left(a^2\sin^2\theta\right)^{3/2} = a^3\,|\sin\theta|^3, $$

not $a^3\sin^3\theta$ — the square root of $\sin^2\theta$ raised to an odd power must be taken as $|\sin\theta|$, since $\theta$ ranges over negative values too and $(a^2\sin^2\theta)^{3/2}\geqslant0$ always. So:

$$ \int_0^{a\cos\theta} r\sqrt{a^2-r^2}\,dr = \frac{a^3}{3}\left(1-|\sin\theta|^3\right). $$

Step 3 · Integrate with respect to $\theta$

$$ I_2 = \frac{a^3}{3}\int_{-\pi/2}^{\pi/2}\left(1-|\sin\theta|^3\right)d\theta = \frac{a^3}{3}\left[\pi - \int_{-\pi/2}^{\pi/2}|\sin\theta|^3\,d\theta\right]. $$

Since $|\sin\theta|^3$ is even, $\displaystyle\int_{-\pi/2}^{\pi/2}|\sin\theta|^3\,d\theta = 2\int_0^{\pi/2}\sin^3\theta\,d\theta = 2\cdot\frac23=\frac43.$

Step 4 · Combine the results

$$ I_2 = \frac{a^3}{3}\left(\pi-\frac43\right) = \frac{\pi a^3}{3}-\frac{4a^3}{9}. $$

Additional Notes

• A tempting shortcut in Part $2^\circ$ is to write $(1-\cos^2\theta)^{3/2}=\sin^3\theta$ directly and then note that $\int_{-\pi/2}^{\pi/2}\sin^3\theta\,d\theta=0$ (an odd function over a symmetric interval), which would give the clean-looking but incorrect answer $\frac{\pi a^3}{3}$. The correct identity is $(1-\cos^2\theta)^{3/2}=(\sin^2\theta)^{3/2}=|\sin\theta|^3$, an even function, and using it gives the genuine answer $\frac{\pi a^3}{3}-\frac{4a^3}{9}$.
• Verified numerically at $a=1$: direct double integration gives $I_2\approx0.60275$, matching $\frac\pi3-\frac49\approx0.60275$ (and not the flawed $\frac\pi3\approx1.0472$).