Using polar coordinates, find the volume of the solid

Using polar coordinates, find the volume of the solid bounded by the paraboloid: $$ 2az=x^2+y^2 $$ and the sphere: $$ x^2+y^2+z^2=3a^2, $$ where $x^2+y^2<2az$.

Step 1 · Find the intersection

Substituting $x^2+y^2=2az$ into the sphere: $2az+z^2=3a^2\Rightarrow z^2+2az-3a^2=0\Rightarrow z=a$ or $z=-3a$. Taking the physically relevant root $z=a$ (consistent with $z>0$ on the paraboloid for $x^2+y^2>0$): $x^2+y^2=2a\cdot a=2a^2$, so the intersection is the circle $r=\sqrt2\,a$ at height $z=a$.

Step 2 · Set up the volume integral

For $0\leqslant r\leqslant\sqrt2\,a$, the solid runs from $z=\dfrac{r^2}{2a}$ (paraboloid) up to $z=\sqrt{3a^2-r^2}$ (upper hemisphere):

$$ V = \int_0^{2\pi}\!\int_0^{\sqrt2\,a}\left[\sqrt{3a^2-r^2}-\frac{r^2}{2a}\right]r\,dr\,d\theta = 2\pi\int_0^{\sqrt2\,a}\left[r\sqrt{3a^2-r^2}-\frac{r^3}{2a}\right]dr. $$

Step 3 · Evaluate the first piece

Substituting $u=3a^2-r^2$: at $r=0$, $u=3a^2$; at $r=\sqrt2\,a$, $u=a^2$:

$$ I_1=\int_0^{\sqrt2\,a}r\sqrt{3a^2-r^2}\,dr = \frac13\left[(3a^2)^{3/2}-(a^2)^{3/2}\right] = \frac13\left[3\sqrt3\,a^3-a^3\right] = \frac{a^3}3\big(3\sqrt3-1\big). $$

Step 4 · Evaluate the second piece

$$ I_2=\int_0^{\sqrt2\,a}\frac{r^3}{2a}\,dr = \frac1{2a}\cdot\frac{(\sqrt2\,a)^4}4 = \frac1{2a}\cdot a^3 = \frac{a^3}2. $$

Step 5 · Combine

$$ V = 2\pi\left[\frac{a^3}3\big(3\sqrt3-1\big)-\frac{a^3}2\right] = 2\pi a^3\left[\frac{2(3\sqrt3-1)-3}6\right] = 2\pi a^3\cdot\frac{6\sqrt3-5}6 = \frac{\pi a^3\big(6\sqrt3-5\big)}3. $$

Additional Notes

• This derivation matches the source exactly.
• Verified numerically for $a=1.1$: direct quadrature of the two-piece integral gives $\approx7.5159$, matching $\dfrac{\pi a^3(6\sqrt3-5)}3\approx7.5159$.