How Do You Calculate Surface Integrals for Vector Fields?

[Reshma]

I have two problems on surface integrals.

1] I have a constant vector \vec v = v_0\hat k. I have to evaluate the flux of this vector field through a curved hemispherical surface defined by x^2 + y^2 + z^2 = r^2, for z>0. The question says use Stoke's theorem.

Stoke's theorem suggests:
\int_s \left(\vec \nabla \times \vec v\right) \cdot d\vec a = \int_p \vec v \cdot d\vec l

But the curl of this vector comes out to be zero . Am I going right? How is the surface integral evaluated?

2] I have a vector field \vec A = y\hat i + z\hat j + x\hat k. I have to find the value of the surface integral:
\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a

The surface S here is a paraboloid defined by:
z = 1 - x^2 - y^2

I evaluated the curl and it comes out to be:
\vec \nabla \times \vec A = -1\left(\hat i + \hat j + \hat k\right)

I need help here on the procedure to evaluate the surface integral.

 

[HallsofIvy]

I have two problems on surface integrals.

1] I have a constant vector \vec v = v_0\hat k. I have to evaluate the flux of this vector field through a curved hemispherical surface defined by x^2 + y^2 + z^2 = r^2, for z>0. The question says use Stoke's theorem.

Stoke's theorem suggests:
\int_s \left(\vec \nabla \times \vec v\right) \cdot d\vec a = \int_p \vec v \cdot d\vec l

But the curl of this vector comes out to be zero . Am I going right? How is the surface integral evaluated?

Not , ! The integral of 0 over any volume is 0 so you have the answer right in front of you! Since you are using Stoke's theorem, you don't need to evaluate the surface integral.

2] I have a vector field \vec A = y\hat i + z\hat j + x\hat k. I have to find the value of the surface integral:
\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a

The surface S here is a paraboloid defined by:
z = 1 - x^2 - y^2

I evaluated the curl and it comes out to be:
\vec \nabla \times \vec A = -1\left(\hat i + \hat j + \hat k\right)

I need help here on the procedure to evaluate the surface integral.

You probably have a formula for d\vec a but here is how I like to think about it: The surface is given by z= 1- x²- y² which is the same as x²+ y²+ z= 1. We can think of that as a "level surface" of the function f(x,y,z)= x²+y²+ z. The gradient of that: 2xi+ 2yj+ k is normal to the surface and, since it is "normalized" to the xy-plane in the sense that the k component is 1, d\vec a= (2xi+ 2yj+ k)dxdy.
Take the dot product of that with -(i+ j+ k) to get the integrand. You don't say over what region of the paraboloid that is to be integrated. If it is over the region above z= 0, then projected into the xy-plane, you have the circle x²+ y²= 1. Integrate over that circle. It will probably be simplest to do it in polar coordinates.

 

[Reshma]

Thank you very, very much!

\left(\vec \nabla \times \vec A\right) \cdot d\vec a = -\left(2x + 2y + 1\right)dxdy

The region of integration is over z>0. Just a little doubt...will the projected region on the xy-plane be a circle even if z>0?

\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a = -\int_0^1 \int_0^1 \left(2x + 2y + 1\right)dxdy

\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a = -\int_0^1 \int_0^1 \left(2xdx + 2ydx + 1dx\right)dy

\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a = -2\int_0^1 (1+y)dy

Evaluation of this gives me: -3