Let $K= \left\{ (x,y,z)\mid x^2+y^2+z\leq 4,z\geq 0 \right\} $. I need to calculate the integral $$\int _K 3(x^2 +y^2).$$
I want to use cylindrical coordinates and calculate $$2\pi\int_{z=0}^4\int_{r=0}^{4-z} 3r^2\cdot rdrdz.$$ But I end up with $6\pi4^4$, while the answer is supposed to be $32\pi$. What have I done wrong? I'm using iterated integration move along the $z$-axis...
Edit: sorry, I just noticed $4-z$ should really be $\sqrt {4-z}$. Everything falls to place after that.
The image of the function $z=g(x,y) = 4-x^2-y^2$ is an upside down paraboloid. Think of turning a ("smoothened") solo cup so that its bottom is face up.
Consider then the intersection of the graph of $g$ with the plane $z = 0$. Then you get the circle $x^2+y^2 = 4$. We know that $0 \leq z \leq 4-r^2$, and now we can describe this circle in polar coordinates as $\{(r, \theta): 0 \leq r \leq 2, 0 \leq \theta \leq 2 \pi\}$. Now we can set up the integral;
$$\int_K 3(x^2+y^2) = \int_{0}^{2\pi} \int_{0}^2 r \int_{0}^{4-r^2} 3r^2 dz \ dr \ d\theta$$