How do I calculate the surface of an integral of the second kind?

Question

Calculate the surface of an integral of the second kind $$\iint_\sigma z^3dxdy $$

$\sigma$ - the outer surface of the plane $x+y+z=10$, located in the first octant $(x \geq 0,y \geq 0,z \geq 0 )$

I need the help

Answer 1

The maximum value for $x$ is $10$, so you can integrate between these limits. For any given $x$, the $y$ varies between $0$ and $10-x$. Write $z$ in terms of $x$ and $y$ and you need to solve $$\int_0^{10} dx\int_0^{10-x} dy(10-x-y)^3$$