Finding mass of sphere with density

burton4
Messages
1
Reaction score
0
http://tgnot.com/triple_integral_sphere.jpg

A solid ball of radius a has density given by (2a-p) where p = sqrt(x^2 + y^2 + z^2). Determine its mass.

a) I think it's properly setup by converting to spherical, still iffy if (2a-p)=(2p-p)=p

b) What did I just find and how do I then get mass?Thanks-a-centillion if you reply, this isn't homework, but is part of a practice midterm.
 
Physics news on Phys.org
(2a-p)=p is only true where a=p. That's on the outside of your sphere. On the inside it's just (2a-p). Integrate that.
 
If you take the integral of (2a-p) it should work.
Remember the mass is the integral of the density so the boundaries are p^2sin(phi)dp(dtheta)(dphi)
The boundary conditions are:
phi: 0... pi
theta: 0...2pi
p: 0...a

So, triple integral(2a-p)p^2sinphi(dpdthetadphi)

Sorry, I don't know how to make symbols on here yet but hope this helps!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top