Finding the mass of a solid, using Spherical Coordinates.

[supermiedos]

Homework Statement

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).

Homework Equations

m = ∫∫∫δdV

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°.

Thanks in advance

 

[HallsofIvy]

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.

 

[supermiedos]

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.

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

 

[LCKurtz]

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

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

 

[supermiedos]

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

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

 

[HallsofIvy]

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.

 

[supermiedos]

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.

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