Given is a Paraboloid delimited as following: $$z_1 = a(x^2 + y^2),\ z_2 = h $$
That's my try for the Volume computation:
First I find the radius of the circle resulting from the intersection between $z_1$ and $z_2$ :
$ z = a(x^2 + y^2) = h \\ \frac{h}{a} = x^2 + y^2 $
so that means that the radius is $ \sqrt{ \frac{h}{a}}$.
Setting up the triple integral:
$$\int_0^{\sqrt{ \frac{h}{a}}} dr \int_0^{2\pi} d \theta \int_h^{a(x^2 + y^2)} dz \cdot 1$$
Now the change of variables:
$ x = r cos \theta \ \ y = r sin \theta$ , where the Jacobian $J = r$.
The integral would then look like this:
$$\int_0^{\sqrt{ \frac{h}{a}}} dr \int_0^{2\pi} d \theta \int_h^{ar^2} dz \cdot r = - \frac{1}{2} \pi \frac{h^2}{a}$$ The negative sign, I assume, is due to fact that at the first integration I integrated from $h$ to $ar^2$ and the other way around.
Is that correct?
Now, how do I compute the surface of the paraboloid ?
The correct set up for the volume in cylindrical coordinates should be
$$\int_0^{2\pi} d \theta \int_0^h \, dz \int_0^{\sqrt{ \frac{z}{a}}}r \,dr $$
For the surface let use
$$S=2\pi \int_0^h f(z)\sqrt{1+[f’(z)]^2}\,dz$$
with $$f(z)=\sqrt{ \frac{z}{a}}$$