Use double integration with polar coordinates to find the volume of the solid body defined by $B=\{(x,y,z)\in\mathbb{R}^3:z\ge 0,x^2+y^2+z^3\le 6\}$ - note the power of $z$ is $3$, not $2$.
My attempt: First I find the region in the $xy$ plane where the curve intersects it - this occurs when $z=0$, that is, $x^2+y^2=6$. So the region in the $xy$ plane is a circle centred at the origin with radius $\sqrt{6}$. As such we have the following integral :$$\int_{0}^{2\pi}\int_{0}^{\sqrt{6}}r(6-r^2)^{\frac{1}{3}}drd\theta.$$
To evaluate this integral, consider first the inner integral: $\int_{0}^{\sqrt{6}}r(6-r^2)^{\frac{1}{3}}dr=\frac{3}{8}\times 6^{\frac{4}{3}}$ [I THINK...]\
Outer integral - we only multiply this result by $\theta$, and the final result would come out as being $\frac{3\pi\times 6^{\frac{4}{3}}}{4}$.
Is this answer correct?