triple integral on cone

Question

Hello everyone I have to calculate $\int\int\int (x^2+y^2+z^2)^\alpha dxdydz$ on the cone $z=\sqrt{x^2 + y^2}$ which has a height of 1 and base circumference $x^2+y^2=1$. $\alpha >0$.

I considered the cylindrical coordinates

$ x= \rho cos\theta; y=\rho sen\theta; z=p$ where $\rho \in[0,1]$ and $\theta \in[0,2\pi]$.

But what about z? With this parameterization, the integral no longer seems to depend on 3 parameters but on two

Answer 1

The question is not clear, but the only way of turning $z=\sqrt{x^2+y^2}$ into a cone with height equal to $1$ and with a base circumference equal to $x^2+y^2=1$ consists in taking the range of $z$ as equal to $[0,1]$ or to $[-1,0]$ (the answer will be the same in both cases).

For instance, if the range of $z$ is $[0,1]$, then, in cylindrical coordinates, your integral becomes$$\int_0^{2\pi}\int_0^1\int_0^z(\rho^2+z^2)^\alpha\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta=\pi\frac{2^{\alpha +1}-1}{2\alpha^2+5\alpha+3}.$$

Answer 2

What do you mean there is no $3$ parameters?.

After change of coordinates, the integral becomes like this:- $$\int_{0}^{1}\int_{0}^{2\pi}\int_{r}^{1}(r^{2}+z^{2})^{\alpha}\,r\,dzd\theta dr$$