How Does the Divergence Theorem Apply to a Vector Field on a Unit Sphere?

[jlmac2001]

I need help evaluating both sides of the divergence theorem if V=xi+yj+zk and the surface S is the sphere x^2+y^2+z^2=1, and so verify the divergence theorem for this case.

Is the divergence theorem the triple integral over V (div V) dxdydz= the double integral over S (V dot normal)dS? If so I would I evaluate it for the above problem?

 

[HallsofIvy]

This was posted under the "calculus" area and I answered there. Please do not double post.

 

[M98Ranger]

Yes, the divergence theorem states that the triple integral over V of the divergence of a vector field V is equal to the double integral over the surface S of the dot product of V and the unit outward normal vector dS. In this case, we have V=xi+yj+zk and S is the sphere x^2+y^2+z^2=1.

To evaluate the left side of the theorem, we first need to find the divergence of V. The divergence of a vector field V = (V1, V2, V3) is given by div V = ∂V1/∂x + ∂V2/∂y + ∂V3/∂z. In this case, we have V1=x, V2=y, and V3=z, so the divergence of V is simply 1+1+1=3.

Next, we need to evaluate the triple integral over V of the divergence of V. Since V is a constant vector field, we can pull it out of the integral and evaluate only the divergence. This gives us ∫∫∫Vdiv V dxdydz = ∫∫∫3 dxdydz = 3∫∫∫dxdydz. Since V is defined over the entire volume V, this integral is equivalent to the volume of V, which in this case is the volume of the unit sphere. Therefore, the left side of the divergence theorem is equal to the volume of the unit sphere.

To evaluate the right side of the theorem, we need to find the dot product of V and the unit outward normal vector dS. The unit outward normal vector for a sphere is given by n = (x, y, z)/√(x^2+y^2+z^2), which in this case is simply n = (x, y, z). Therefore, the dot product V dot n is equal to x^2+y^2+z^2, which is equal to 1 on the surface S.

Next, we need to evaluate the double integral over S of V dot n dS. Since V dot n is a constant (equal to 1) on the surface S, we can pull it out of the integral and evaluate only the surface area dS. Since S is a sphere with radius 1, the surface area dS is given by dS = r^2sinθd