# Finding the mass of a solid, using Spherical Coordinates.

1. Jul 30, 2015

### supermiedos

1. The problem statement, all variables and given/known data
Find the mass of the solid bounded from above z = √(25 - x2-y2) and below from z = 4, if its density is δ = k(x^2 + y^2 + z^2)^(-1/2).

2. Relevant equations
m = ∫∫∫δdV

3. The attempt at a solution

The plane z = 4 is transformed into ρcosφ = 4, that is, ρ = 4secφ. And x^2 + y^2 + z^2 = 25 is ρ = 5. θ goes from 0 to 2π. But I'm struggling finding the limits for φ (the azimuth). I think φ must go from 0 to π/2, but I can't get the correct answer (which is kπ). The book suggests that I must express the upper limit for φ as an inverse cosine, but why is that? From the figure I can see φ goes from 0 to 90°.

2. Jul 30, 2015

### HallsofIvy

Staff Emeritus
Yes, $\rho$, the radial variable, goes from 0 to 5 and $\theta$, the "longitude" goes from 0 to $2\pi$. $\phi$, the "co-latitude" goes from 0 to arcsin(1/5) since line making angle $\phi$ with the z-axis, to a point on the hemi-sphere with z= 1, is the hypotenuse of a right triangle with "opposite side" 1.

3. Jul 30, 2015

### supermiedos

Wait, ρ goes from 0 to 5? I tought ρ went from 4secφ to 5. And why do you use z = 1?

4. Jul 30, 2015

### LCKurtz

You are correct for $\rho$ and $\phi$ from 0 to $\arcsin(\frac 3 5)$ or $\arccos(\frac 4 5)$.

5. Jul 30, 2015

### supermiedos

Thanks both of you for your help. The result it's correct :)

6. Jul 31, 2015

### HallsofIvy

Staff Emeritus
I used z= 1 because I misread your post, thinking the base was 4 places below the top, at z= 1, rather than at z= 4. Of course, you should use z= 4, not z= 1.

7. Jul 31, 2015

### supermiedos

Thank you, and thanks for the idea of the triangle.