How can I compute the volume integral of a parameterized set?

Question

This is really odd to me. How can I compute the volume of this cup?

Answer 1

Parametrized surface: $x = r \cos\varphi, \ y = r \sin \varphi - (1-\sqrt{4-z})^2, z$

$r \geq 0, 0 \leq \varphi \leq 2\pi, 3\sqrt r \leq z \leq 3$

As we can see from parametrization that for any given $z$, we have a circle of radius $r$ with center at $(0, - (1-\sqrt{4-z})^2, z)$.

Now $z \geq 3 \sqrt r \implies r \leq \frac{z^2}{9}$.

So using disk method, the volume can be found as

$\displaystyle \int_0^{2\pi} \int_0^{3} \int_0^{z^2/9} r \ dr \ dz \ d\varphi = \frac{3 \pi}{5}$

Using shell method,

$\displaystyle \int_0^{2\pi} \int_0^{1} \int_{3 \sqrt r}^3 r \ dz \ dr \ d\varphi = \frac{3 \pi}{5}$

(the upper bound of $r$ comes from $z = 3 \sqrt r = 3 \ $ or $r = 1)$

As Martin R said, this is same as the volume of a cone of same radius and height (Cavalieri's Principle)