Using polar coordinates, find the volume of the solid

Using polar coordinates, find the volume of the solid bounded by the paraboloid: $$ z=3-x^2-y^2 $$ and the plane $z=0$.

Step 1 · Find the base disk

Setting $z=0$: $x^2+y^2=3$, so the base region is $D=\{x^2+y^2\leqslant3\}$.

Step 2 · Convert to polar coordinates

With $x=r\cos\theta,\,y=r\sin\theta$ (Jacobian $r$), $z=3-r^2$, and:

$$ V = \int_0^{2\pi}\!\int_0^{\sqrt3}(3-r^2)\,r\,dr\,d\theta = \int_0^{2\pi}\!\int_0^{\sqrt3}\big(3r-r^3\big)\,dr\,d\theta. $$

Step 3 · Evaluate the inner integral

$$ \int_0^{\sqrt3}(3r-r^3)\,dr = \left[\frac{3r^2}2-\frac{r^4}4\right]_0^{\sqrt3} = \frac{9}2-\frac94 = \frac94. $$

Step 4 · Evaluate the outer integral

$$ V = \int_0^{2\pi}\frac94\,d\theta = \frac94\cdot2\pi = \frac{9\pi}2. $$

Additional Notes

• This derivation matches the source exactly. As a sanity check: the volume of a paraboloid of height $h$ over a disk of radius $\rho$ is exactly half the volume of the circumscribing cylinder $\pi\rho^2h$; here $h=3,\ \rho=\sqrt3$, giving $\frac12\pi(3)(3)=\frac{9\pi}2$, matching.
• Verified symbolically: direct evaluation of the nested integral gives exactly $\dfrac{9\pi}2$.