Using a double integral, find the volume of the oblate

Using a double integral, find the volume of the oblate ellipsoid of revolution $\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2}+\dfrac{z^2}{b^2}=1$.

Explanation of Method

By symmetry about the $xy$-plane, the volume is twice the volume under $z=b\sqrt{1-\frac{x^2+y^2}{a^2}}$ over the disk $x^2+y^2\le a^2$. Polar coordinates reduce this to a single elementary radial integral.

Express $z$ from the ellipsoid equation and identify the base disk.
Convert to polar coordinates.
Evaluate via $u=1-r^2/a^2$.

Step 1 · Setup

$z=b\sqrt{1-\dfrac{x^2+y^2}{a^2}}$ over $D:x^2+y^2\le a^2$, and $V=2\iint_Dz\,dx\,dy$.

Step 2 · Polar coordinates

$$ V = 2\int_0^{2\pi}\int_0^ab\sqrt{1-\frac{r^2}{a^2}}\,r\,dr\,d\theta = 4\pi b\int_0^ar\sqrt{1-\frac{r^2}{a^2}}\,dr. $$

Step 3 · Evaluate

With $u=1-\dfrac{r^2}{a^2}$:

$$ V = 2\pi a^2b\int_0^1u^{1/2}\,du = 2\pi a^2b\cdot\frac23 = \frac{4\pi a^2b}3. $$

Additional Notes

This derivation matches the source exactly, and is consistent with the general ellipsoid formula $V=\frac43\pi abc$ with two equal semi-axes $a$ and the third $b$.
Verified numerically: direct double integration (e.g. $a=1.7,\,b=2.3$) gives $\approx27.842888$, matching $\dfrac43\pi a^2b$ exactly.

Result

The volume is:

$$ V = \frac43\pi a^2b. $$