Iterated integral in spherical coordinates used to find mass

[LoffieZA]

Attached is a question asking for the iterated integral in order to find the mass for the bottom section of the sphere. My working is attached but I am not feeling very confident especially when it comes to the boundaries of phi and P. The actual mass is not required, only the integral which would find the mass. I also took into consideration the factor p^2 sin(phi)

I would appreciate it dearly if someone could check my working for me and possibly give me a tip on how to approach such a problem.

 

[LCKurtz]

It looks good except for your lower limit for [MATH]\phi[/MATH]. Draw a [MATH]zy[/MATH] cross section (a circle with a line at [MATH]z=-2[/MATH]), and draw the line from the origin to the intersection of the line and circle. You should be able to read the min value of [MATH]\phi[/MATH] from that picture. What do you get?

 

[LoffieZA]

It looks good except for your lower limit for [MATH]\phi[/MATH]. Draw a [MATH]zy[/MATH] cross section (a circle with a line at [MATH]z=-2[/MATH]), and draw the line from the origin to the intersection of the line and circle. You should be able to read the min value of [MATH]\phi[/MATH] from that picture. What do you get?

Thanks for the quick response. I did what you suggested and my work is attached. It does seem unusual though because the point (sqrt(5), -2) does not fall onto one of the usual unit circle angles e.g pi/3. Is there something that I am still missing? In radians the angle would be about 2.3 rad.

 

[LCKurtz]

You know [MATH]z = -2[/MATH] and [MATH]r=3[/MATH] in your little triangle. Why not just leave the angle as an arccos, no decimals needed.

 

[LoffieZA]

You know [MATH]z = -2[/MATH] and [MATH]r=3[/MATH] in your little triangle. Why not just leave the angle as an arccos, no decimals needed.

O ok, I see what you mean. That would make the lower boundary for phi be (pi/2) + arcsin(2/3). Correct?

 

[LCKurtz]

That is correct but unnecessarily complicated. Why not just [MATH]\arccos (-\frac 2 3)[/MATH]?

 

[LoffieZA]

oh true. Makes much more sense. Thanks so much for you guidance and patience. I appreciate it.