Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D \sqrt{4-\left(\frac xa\right)^2-\left(\frac yb\right)^2}\,dx\,dy, $$ where $D$ is the first-quadrant region bounded by the ellipses $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and $\dfrac{x^2}{(2a)^2}+\dfrac{y^2}{(2b)^2}=1$.

Step 1 · Elliptical substitution

Let $x=au$, $y=bv$; then $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=u^2+v^2$, and $dx\,dy=ab\,du\,dv$. The integrand becomes $\sqrt{4-u^2-v^2}$. The inner ellipse becomes $u^2+v^2=1$ and the outer becomes $u^2+v^2=4$, so:

$$ I = ab\iint_{1\leqslant u^2+v^2\leqslant4,\,u,v\geqslant0} \sqrt{4-u^2-v^2}\,du\,dv. $$

Step 2 · Convert to polar coordinates

With $u=r\cos\theta$, $v=r\sin\theta$: $1\leqslant r\leqslant2$, $0\leqslant\theta\leqslant\frac{\pi}{2}$:

$$ I = ab\int_0^{\pi/2}\int_1^2 \sqrt{4-r^2}\,r\,dr\,d\theta. $$

Step 3 · Integrate with respect to $r$

With $w=4-r^2$, $dw=-2r\,dr$; when $r=1$, $w=3$, and when $r=2$, $w=0$:

$$ \int_1^2 r\sqrt{4-r^2}\,dr = \frac12\int_0^3\sqrt w\,dw = \frac12\left[\frac{2w^{3/2}}{3}\right]_0^3 = \frac{3^{3/2}}{3} = \sqrt3. $$

Step 4 · Integrate with respect to $\theta$

$$ I = ab\int_0^{\pi/2}\sqrt3\,d\theta = ab\cdot\sqrt3\cdot\frac{\pi}{2} = \frac{\pi ab\sqrt3}{2}. $$

Additional Notes

• Verified numerically (with $a=2$, $b=3$): direct double integration gives $I\approx16.324$, matching $\dfrac{\pi ab\sqrt3}{2}\approx16.324$.