Help Evaluate ∫∫F∙dS on Surface Integrals

[bodensee9]

Please Help! Surface integrals

I am wondering if someone can help me with the following? I am asked to evaluate ∫∫F∙dS where F(x,y,z) = z^2xi + (1/3y^3 +tanz)j + (x^2z+y^2)k and S is the top half of the sphere x^2+y^2+z^2 = 1.

∫∫F∙dS = ∫∫∫divFdV. Here, div F = x^2+y^2+z^2. I know that S is not a closed surface and so you would need to evaluate S as the difference between 2 surfaces, S1 as the closed surface that is the top half of the sphere and S2 as the disk that is x^2+y^2≤1 where the orientation is downwards. So, evaluating div F over the whole top half of the sphere I got 2pi/5.

But I am wondering how I would evaluate ∫∫∫x^2+y^2+z^2 over the disk x^2+y^2≤1? Could I convert to spherical coordinates and I could simplify the expression of the integrant to ρ^4sinφ, but I am wondering what φ would be in this instance? Thanks so much!

 

[HallsofIvy]

I am wondering if someone can help me with the following? I am asked to evaluate ∫∫F∙dS where F(x,y,z) = z^2xi + (1/3y^3 +tanz)j + (x^2z+y^2)k and S is the top half of the sphere x^2+y^2+z^2 = 1.

∫∫F∙dS = ∫∫∫divFdV. Here, div F = x^2+y^2+z^2. I know that S is not a closed surface and so you would need to evaluate S as the difference between 2 surfaces, S1 as the closed surface that is the top half of the sphere and S2 as the disk that is x^2+y^2≤1 where the orientation is downwards. So, evaluating div F over the whole top half of the sphere I got 2pi/5.

But I am wondering how I would evaluate ∫∫∫x^2+y^2+z^2 over the disk x^2+y^2≤1? Could I convert to spherical coordinates and I could simplify the expression of the integrant to ρ^4sinφ, but I am wondering what φ would be in this instance? Thanks so much!

You wouldn't integrate a volume integral over a surface!

$\int\int\int (x^2+ y^2+ z^2)dzdydx$ is integrated over the volume- the half-ball- not the surface. If you are using cartesian coordinates, integrate with x from -1 to 1, y from $-\sqrt{1- x^2}$ to $\sqrt{1- x^2}$, z from $-\sqrt{1- x^2- y^2}$ to $\sqrt{1- x^2- y^2}$. It's simpler in cylindrical coordinates: integrate with r from 0 to 1, $\theta$ from 0 to $\2 pi$, z from 0 to $\sqrt{1- r^2}$. And, of course, it's much simpler in spherical coordinates: integrate with r from 0 to 1, $\theta$ from 0 to $2\pi$, $\phi$ from 0 to $\pi/2$.