move the integral
$\int_0^1 \int_0^{\sqrt{1 - x^2}} \int_0^{\sqrt{1 - x^2 - y^2}} \sqrt{x^2 + y^2 + z^2}\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac{\pi}{8}$
to spherical coordinates integral.
i dont know how to find the limits of the integral when moving to $\theta, \ r , \phi $
can you please guide me with finding the limits because i always get a mistakes there
my trial :
reading the $\mathrm dz$ we see that $ 0 \leq z^2 \leq 1-x^2-y^2 $ or $ 0\leq x^2 + y^2 + z^2 \leq 1 $
reading $\mathrm dy$ we see that $ 0 \leq y^2 \leq \sqrt{1-x^2} $ or $ 0\leq x^2 + y^2 \leq 1$
reading $dx$ we see that $ 0 \leq x \leq 1$
so the spherical limits :
$ 0\leq r^2 \leq 1$
$ 0 \leq r^2\sin(\theta)^2 \leq 1$
$ 0 \leq r\cos(\theta)\sin(\phi) \leq 1$
how do i continue from here ? i think i made a mistake with $\phi$ because i didn't get the proper value for the integral after calculating in the spherical co-ordinates.
It follows from your computations that your integral is equal to$$\int_0^1\int_0^{\frac\pi2}\int_0^{\frac\pi2}r^3\sin(\phi)\,\mathrm d\theta\,\mathrm d\phi\,\mathrm dr.$$But\begin{align}\int_0^1\int_0^{\frac\pi2}\int_0^{\frac\pi2}r^3\sin(\phi)\,\mathrm d\theta\,\mathrm d\phi\,\mathrm dr&=\left(\int_0^1r^3\,\mathrm dr\right)\left(\int_0^{\frac\pi2}\sin(\phi)\,\mathrm d\phi\right)\left(\int_0^{\frac\pi2}\,\mathrm d\theta\right)\\&=\frac14\times1\times\frac\pi2\\&=\frac\pi8.\end{align}