Triple Integral in Spherical Coordinates.

Question

$\newcommand{\de}{\operatorname{d}}$A little stuck on this one.

$$\iiint_V ye^{-(x^2+y^2+z^2)^2}\,{\rm d} V$$ Use Spherical Coordinates to evaluate where V is the solid that lies between y=0 and the hemisphere $x^2+y^2+z^2=1$ in the right half space $(y>0)$.

So my bounds will be $0\le r\le1$, $0\le \theta\le\pi$, $0\le \phi\le\pi$.

The integral will become $$\int_0^1\int_0^\pi\int_0^\pi y\,e^{-(r^2)^2}r^2\sin(\phi)\,{\rm d} \phi \,{\rm d} \theta \,{\rm d} r$$

I'm not too sure how to solve it out from this point :(

In advance, thanks for your help!

Answer 1

As has been commented, you need to substitute $y=r\sin(\theta)\sin(\phi)$ into your final equation. You will then have

$$\int_0^1\int_0^\pi\int_0^\pi r^3\sin(\theta)\sin^2(\phi)e^{-r^4}d\phi\;d\theta\;dr=\int_0^1r^3e^{-r^4}dr\int_0^\pi\sin(\theta)d\theta\int_0^\pi\sin^2(\phi)d\phi$$ which can be easily integrated, realising that $\sin^2(\phi)=\frac12(1-\cos(2x))$.

Answer 2

You forgot to substitute $y=r\sin\theta\sin\phi$

Try:$$\iiint_V ye^{-(x^2+y^2+z^2)^2}\de V =\int_0^1\int_0^\pi\int_0^\pi e^{-r^4}r^3\sin(\theta)\sin^2(\phi)\de \phi \de \theta \de r$$

Answer 3

Remember to substitute $y=r\sin\theta\sin\phi$. (Likewise you would use $x=r\cos\theta\sin\phi, z=r\cos\phi$, if they were needed.)

Then the integral can be rewritten as follows:

$$\int_0^1\int_0^\pi\int_0^\pi r^3\sin \theta \sin^2 \phi \,e^{-r^4} \de \phi \de \theta \de r \\ = \int_0^1 r^3 e^{r^4}\de r \times \int_0^\pi \sin \theta \de \theta \times \int_0^\pi \sin^2 \phi \de \phi$$

Can you take it from here?