Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D \sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,dx\,dy, $$ where $D$ is the region $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\leqslant1$.

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 the Jacobian is:

$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}a&0\\0&b\end{vmatrix} = ab, \qquad dx\,dy = ab\,du\,dv. $$

The region becomes the unit disk $u^2+v^2\leqslant1$:

$$ I = ab\iint_{u^2+v^2\leqslant1} \sqrt{1-u^2-v^2}\,du\,dv. $$

Step 2 · Convert to polar coordinates

With $u=r\cos\theta$, $v=r\sin\theta$, $0\leqslant r\leqslant1$, $0\leqslant\theta\leqslant2\pi$:

$$ I = ab\int_0^{2\pi}\int_0^1 \sqrt{1-r^2}\,r\,dr\,d\theta. $$

Step 3 · Integrate with respect to $r$

With $w=1-r^2$, $dw=-2r\,dr$:

$$ \int_0^1 r\sqrt{1-r^2}\,dr = \frac12\int_0^1\sqrt w\,dw = \frac12\left[\frac{2w^{3/2}}{3}\right]_0^1 = \frac13. $$

Step 4 · Integrate with respect to $\theta$

$$ I = ab\int_0^{2\pi}\frac13\,d\theta = ab\cdot2\pi\cdot\frac13 = \frac{2\pi ab}{3}. $$

Additional Notes

• This is the volume under the upper half of the ellipsoid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+z^2=1$. Since the full ellipsoid with semi-axes $a,b,1$ has volume $\dfrac{4\pi ab}{3}$, the upper half has exactly half that volume, $\dfrac{2\pi ab}{3}$ — matching the result above.