$$\int \int \int _V\:xyzdxdydz$$ with V:$x^2+y^2+z^2=2$ and $z=x^2+y^2$
I first did the drawing, $x^2+y^2+z^2=2$ being a sphere with radius $\sqrt{2}$ and $z=x^2+y^2$ its a paraboloid.
I don't know how to follow up now, how should I do it using cylindrical coordinates or sphere or what I have tried is doing the intersection between the sphere and paraboloid so I got that they intersect in the circle $x^2+y^2=1$ thus:
$x^2+y^2\le z\le \sqrt{2-x^2-y^2}$
I just want to know how to do the parametrics correct using cylindrica or sphere coordinates or the way I tried, no need to evaluate the integral!
Best Answer
You have been asked to evaluate $\displaystyle \iiint_V\:xyz \ dx \ dy \ dz$ over a region which is bound by $z = x^2 + y^2$ and $x^2 + y^2 + z^2 = 2$.
What you need to observe is that in the region, $z \geq 0$ and the region is symmetric to $XZ$ and $YZ$ planes and the function $xyz$ is an odd function wrt $x$ and $y$. It is clear that $xyz$ is positive in two octants and negative in two octants. So the integral will simply be zero.
In cylindrical coordinates, $x = r \cos \theta, y = r \sin\theta, z = z$ so equations of surfaces can be written as,
$z = r^2, r^2 + z^2 = 2$.
That gives you bounds of $z$. At the intersection of paraboloid and sphere,
$r^2 + r^4 = 2 \implies r = 1$
So integral should be,
$\displaystyle \int_0^{2\pi} \int_0^1 \int_{r^2}^{\sqrt{2-r^2}} r^3 z \cos \theta \sin \theta \ dz \ dr \ d\theta$
As I said in the beginning, the integral is zero and we know that. But it depends on what the exercise is about.