calculating the solid volume limited by the $xy$ plane and the surfaces $z=\frac{x^2}{2p}+\frac{y^2}{2q}$ and $x^2+y^2=a^2$ where $p,q,a ∈ R^+$ help me with this excercises
calculating the solid volume limited by the $xy$ plane and the surfaces
$$z=\frac{x^2}{2p}+\frac{y^2}{2q}$$
$$x^2+y^2=a^2$$
where $p,q,a ∈ R^+$
i dont know how to start this excercises, help !!!
 A: Start by thinking about what each surface looks like:
$xy$ plane is pretty straightforward. It is $z=0$.
$x^2+y^2=a^2$ is 2D would describe a circle with radius $a$ so in 3D it describes a vertical cylinder with radius $a$.
$z=\frac{x^2}{2p}+\frac{y^2}{2p}$ is harder to visualize by we can tell that $z$ here is a function of $x$ and $y$ and that is it always non-negative so it must be above (or equal to) the $xy$ plane. So your integral can be written as:
$$\iint_{x^2+y^2=a^2}\frac{x^2}{2p}+\frac{y^2}{2p}dA$$
To evaluate it:
From the shape using polar coordinates is going to make it easy. So rewrite $x=r\cos\theta$ and $y=r\sin\theta$. The integral becomes:
$$\int_0^{2\pi}\int_0^a \left(\frac{(r\cos\theta)^2}{2p}+\frac{(r\sin\theta)^2}{2p}\right)r\,dr\,d\theta$$
$$=\frac{2\pi}{2p}\int_0^a r^3\left(\cos^2\theta+\sin^2\theta\right)dr$$
$$=\frac{\pi}{p}\int_0^a r^4\,dr$$
$$=\frac{\pi}{p}\times\frac{a^5}{5}$$
$$=\frac{\pi a^5}{5p}\ units^3$$
A: You can use the polar form: $$x=r\cos(t),~y=r\sin(t),~ 0\leq t<2\pi$$ So we have $$z=\frac{x^2}{2p}+\frac{y^2}{2q}\longrightarrow z=r^2\left(\frac{\cos^2(t)}{2p}+\frac{\sin^2(t)}{2q}\right)$$ and the new ranges are $0\leq t<2\pi,~~ 0\leq r\leq a$. Don't forget to use Jacobian in the new integral.
