You have ∫∫∫ r^2 sinθ cosØ dA
so change the integration limits
online class and I don't know what to do so I posts. Any help is great.Thank you
Convert the integral [0,1]∫ [0,√(1-x^2 -y^2)]∫𝑧√(x^2 +y^2 +z^2)dz dy dx into anequivalent integral in spherical coordinates and evaluate the integral.
3 answers
How did you get r^2 sinθ cosØ dA. Can you show a some sample of how you would set them up. I don't understand with just ∫∫∫ r^2 sinθ cosØ dA for the question.
Convert the integral [0,1]∫ [0, √(1-x^2)∫ [0,√(1-x^2 -y^2)]∫𝑧√(x^2 +y^2 +z^2)dz dy dx into an equivalent integral in spherical coordinates and evaluate the integral.
Convert the integral [0,1]∫ [0, √(1-x^2)∫ [0,√(1-x^2 -y^2)]∫𝑧√(x^2 +y^2 +z^2)dz dy dx into an equivalent integral in spherical coordinates and evaluate the integral.
Hmmm. I did mess up on that one; I didn't refer to my vector analysis text first. However, you only have a double integral, and 3 variables. I expect that the limits for y are [1,√(1-x^2)].
This one is similar to the polar coordinates one. The volume element
dv = dz dy dx becomes dv = r^2 sinθ dr dθ dØ
If that isn't clear, google volume element in spherical coordinates to see how this is done. Your class materials surely also derive the transformation.
Now, with z = r cosθ
the integrand z √(x^2 +y^2 +z^2) = r cosθ * r = r^2 cosθ and so the integral becomes ∫∫∫ r^2 cosθ * r^2 sinθ dr dθ dØ
and the region appears to be the 1st octant of the sphere of radius 1, so the limits on θ and Ø should be pretty easy. See what you can come up with.
This one is similar to the polar coordinates one. The volume element
dv = dz dy dx becomes dv = r^2 sinθ dr dθ dØ
If that isn't clear, google volume element in spherical coordinates to see how this is done. Your class materials surely also derive the transformation.
Now, with z = r cosθ
the integrand z √(x^2 +y^2 +z^2) = r cosθ * r = r^2 cosθ and so the integral becomes ∫∫∫ r^2 cosθ * r^2 sinθ dr dθ dØ
and the region appears to be the 1st octant of the sphere of radius 1, so the limits on θ and Ø should be pretty easy. See what you can come up with.