1. ## Vector Calculus

A solid is defined by a surface:
x^2 + y^2 + z^2 ≤ b^2

Whereby, 0 ≤ x, 0 ≤ y, 0≤ z

The density varies in proportion to the distance from the centre.

Hence deduce the mass of the solid.

I assume i use sphereical polar coordinates, but not entirely sure where i go beyond that.

Any help is appreciated!

2. In general, if you want to compute the mass of an object occupying volume $V$ with density $\rho,$ then you must compute the integral

$M=\iiint_{V}\rho\,dV.$

Spherical coordinates are definitely the way to go, although you could probably work with cylindrical as well. As I see it, you need to assemble three pieces: the limits for each integral, the density, and the volume differential. What are each of those?

3. You could calculate $\int_0^bp(r)S(r)\,dr$ where p(r) is the density at distance r from the center and S(r) is 1/8 of the area of a sphere of radius r.

4. ok the method i've gone for (which i've been told works)

$M = \iiint \rho (r) f(r,\varphi, \theta) |J| dr d\varphi d\theta
$

Where J is the Jacobian (which i deduced to be ${-r}^{2 }sin(\theta)$

Note $\rho (r) = rk$ where k is a constant

For some reason (which i do not fully understand), $f(r,\varphi, \theta) = 1$. Hence the integral simplifies.

However, im struggling to compute the limits of integration. I know 0 ≤ x, 0 ≤ y, 0≤ z
So we're concentrating in the first octant, but how do i deduce the integrals?
Is it just r between 0 and r, $\varphi$ between 0 and $\frac{\pi }{2 }$ and the same for $\theta$? Or is $r = {a}$

Hope you can clear my confusion

5. It would seem $\theta$ has limits $\pi ...0$

Which gets me what seems to be a reasonable answer of:
$\frac{k{a}^{4}\pi}{2}$

Can anyone explain this, if this is correct, i dno not know how to choose the correct limits

6. The first octant, which is the octant over which you're integrating your sphere, has limits of the polar angle $\theta\in[0,\pi/2],$ and the azimuthal angle $\varphi\in[0,\pi/2].$

I don't think your Jacobian is correct. The element of volume is positive, not negative. You have the correct magnitude.