How Do You Solve the Third Integral in a Triple Integral Problem?

[acedeno]

Homework Statement

the limits of integration are s[0,R] p[0,2π] z[0,L] respectively

kp∫∫∫s/sqrt(s^2+(d-z)^2)dsdpdz

Homework Equations

The Attempt at a Solution

You can double check my work but I'm pretty sure I got the first to integrations fine (with respect to s and p), I'm just not sure about how to do the third integral with respect to z

so basically I'm stuck here:

2πkp∫[sqrt(R^2+(d-z)^2) - (d-z)]dz

 

[mighty2000]

Hello,

Thank you for sharing your work with us. It looks like you have correctly set up the first two integrations with respect to s and p. To solve the third integral with respect to z, you can use the substitution u = d-z. This will allow you to rewrite the integral as:

2πkp∫[sqrt(R^2+u^2) - u]du

This integral can be solved using trigonometric substitution or by using the substitution v = R^2+u^2. Once you have solved the integral, don't forget to substitute back in for u = d-z and include the limits of integration for z, which are 0 and L.

I hope this helps. Good luck with your calculations!