Fairly simple question, we have the paraboloid $z=a(x^2+y^2)$ and the plane $z=h$. $a,h >0$
Find the volume of the region bounded by the plane and the paraboloid.
What I did:
It is clear to see $a(x^2+y^2) \leq z \leq h$ from the way the question is written.
The hard part which im not sure of is the limits of $x,y$.
$a(x^2+y^2) = ax^2+ay^2 \leq h$ implies: $- \sqrt{\frac{h-ax^2}{a}}\leq y \leq \sqrt{\frac{h-ax^2}{a}}$
the minimal value of $y^2$ is zero. and if that happens, then $ax^2 \leq h$, which implies $-\sqrt{\frac{h}{a}} \leq x \leq \sqrt{\frac{h}{a}}$
So the volume we are looking for is in my opinion:
$$\int_{-\sqrt{\frac{h}{a}}}^{\sqrt{\frac{h}{a}}} \int_{- \sqrt{\frac{h-ax^2}{a}}}^{ \sqrt{\frac{h-ax^2}{a}}} \int_{a(x^2+y^2)}^{h}dzdydx$$
Is this correct?
Best Answer
Using Cylindrical coordinates. Starting: $ax^2 + ay^2 = h \to x^2 + y^2 = r_0^2$ with $r_0 = \sqrt{\dfrac{h}{a}}$. So set $x = r\cos\theta$, and $y = r\sin\theta$, then we integrate:
$z: ar^2 \to h$
$r: 0 \to \sqrt{\dfrac{h}{a}}$
$\theta: 0 \to 2\pi$. Thus the set up is:
$V = \displaystyle \int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{h}{a}}} \int_{ar^2}^h rdzdrd\theta$