# Homework Help: 2 Surface integral problems

1. Feb 17, 2006

### 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 :yuck:. 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.

Last edited: Feb 17, 2006
2. Feb 17, 2006

### HallsofIvy

Not :yuck:, ! 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.

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.

Last edited by a moderator: Feb 17, 2006
3. Feb 17, 2006

### 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

Last edited: Feb 17, 2006