Physics help?

A particle has the wave function:
Phi(r) = N*(e^(−a r))
where N and a are constants

I worked out previously that N=sqrt(a^3/pi)

The question is: find the expectation value of x

I also know x = rsin(theta)cos(phi) (where phi is a little phi not a capital phi as before, i.e. they different)
y = rsin(theta)sin(phi)
z=rcos(theta)
and dV = r^2dr sin(theta)d(theta)d(phi)

My attempt:
I assume this is no different from usual ways of finding <x> in that you use phi* (x) phi, I tried using a triple integral and substituting x=rsin(theta)cos(phi) in but couldn't make it work, so perhaps this idea is wrong?? (I ended up integrating over 0), anyway, help would be GREATLY APPRECIATED

This seems like a problem with the boundary conditions you've chosen for the spherical coordinates.

I'm thinking r: [0,r], theta: [0, 2pi], small phi: [0, pi], but it's been a while since I've done elementary quantum physics.

Sorry, should have given the boundary coinditions to begin with:

r: [0, infinty)
Theta: [0, pi]
Small phi [0, 2pi)

So when I integrate over small phi, I get Sin(2 pi), which is zero, and hence buggers everything up...

Check your textbook to see that you're choosing the boundary conditions properly. You are calculating the expectation as integral(x*phistar*phi), right?

Yeah, and I'm subbing in rsin(theta)cos(phi)for the x.

Basically, I did three pages of a triple integral, and when I got to integrating in terms of small phi, between 0 and 2pi, the term Cos (small phi)I got zero, which leads me to thinking that the triple integral may not be the way forward.

Maybe the answer really is 0?

You have a spherically symmetric wave function -> all its position/momentum expectation values are 0.

integral(x phi* phi) x₀ -> x₀ + dx =
- integral(x phi* phi) -x₀ -> -x₀ -dx

@6

You're right.

I wasw just expecting the question to be harder than it actually was... I have a fear of QM or something!