Find the volume of the solid bounded by the paraboloid
Find the volume of the solid bounded by the paraboloid: $$ x^2+y^2=2az $$ and the curve: $$ \big(x^2+y^2\big)^2+2a^2xy=0,\qquad z=0. $$
Explanation of Method
In polar coordinates, the paraboloid gives $z=\dfrac{r^2}{2a}$, and the boundary curve reduces to $r^2=-a^2\sin2\theta$ — a two-petaled rose, real only where $\sin2\theta\leqslant0$, i.e. for $\theta\in\left[\frac\pi2,\pi\right]\cup\left[\frac{3\pi}2,2\pi\right]$.
- Convert both surfaces to polar coordinates.
- Set up the volume integral over both petals.
- Evaluate, using symmetry between the two petals.
Step 1 · Polar form
Paraboloid: $r^2=2az\Rightarrow z=\dfrac{r^2}{2a}$. Boundary: $r^4+a^2r^2\sin2\theta=0\Rightarrow r^2=-a^2\sin2\theta$ (for $r\neq0$), real where $\sin2\theta\leqslant0$: $\theta\in\left[\frac\pi2,\pi\right]\cup\left[\frac{3\pi}2,2\pi\right]$, giving $r=a\sqrt{-\sin2\theta}$.
Step 2 · Set up the volume integral
By symmetry, the two petal-integrals are equal:
$$ V = 2\int_{\pi/2}^\pi\int_0^{a\sqrt{-\sin2\theta}}\frac{r^2}{2a}\cdot r\,dr\,d\theta = \frac1a\int_{\pi/2}^\pi\int_0^{a\sqrt{-\sin2\theta}}r^3\,dr\,d\theta. $$
Step 3 · Evaluate the inner integral
$$ \int_0^{a\sqrt{-\sin2\theta}}r^3\,dr = \frac{a^4\sin^22\theta}4. $$
Step 4 · Evaluate the outer integral
$$ V = \frac{a^3}4\int_{\pi/2}^\pi\sin^22\theta\,d\theta = \frac{a^3}4\cdot\frac12\left[\theta-\frac{\sin4\theta}4\right]_{\pi/2}^\pi = \frac{a^3}4\cdot\frac12\cdot\frac\pi2 = \frac{a^3}4\cdot\frac\pi4 = \frac{\pi a^3}{16}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: direct evaluation over both petals gives exactly $\dfrac{\pi a^3}{16}$.
Result
The volume is: