# Homework Help: Stoke's Theorem

1. Feb 3, 2004

### mindcircus

Any help would be MUCH appreciated!

Vector A=yi+zj+xk
Surface S is defined by z=1-x^2-y^2 where z is greater or equal to 0.
Find the value of the surface integral of (grad x A)*dA
(The “x” is the cross product.)

I know this uses Stoke’s Theorem: line integral of the closed curve of A*dS = surface integral (grad x A)*n*dA. So, the line integral of vector A around the closed curve is equal to the curl of A over the surface defined by the curve.

I’m fairly sure I’ve set everything up correctly. I have to find (gradxA). Then I find the unit normal vector by taking the gradient of the surface and dividing it by its magnitude. Then I find dS by making it (dx*dy)/(the magnitude of n*k). Then I do the dot product of my (gradxA) vector and my unit normal vector. Multiply by the magnitude of n*k, then the double integral is set up. I think I have to switch to polar coordinates, at the end.

I figured out (grad x A) by using the determinant. It's -i+j-k.
Now I need to find the normal of the given surface. For this, I take the gradient of the surface. I rearranged the surface so that
x^2+y^2+z=1. The gradient is 2xi+2yj+k. I need the unit vector of the normal, so I divide by the magnitude. This is where I get messed up, and unsure if I'm doing this right. I get the square root of (4x^2+4y^2+1). There's no way this is right, because almost nothing will simplify...

I hope I'm not completely wrong. Thanks!

2. Feb 3, 2004

### HallsofIvy

NO, the line integral of vector A around the closed curve is equal to the integral of curl of A over the surface defined by the curve.
That's probably what you meant, but let's be precise.

I do not get -i+j-k for curl v, I get -i- j- k (and please don't write "grad x v". "grad" is specifically del dot something. You mean "del cross v" or simply "curl v".)

The simplest way to find n*dA is to think of z=1-x^2-y^2 as x^2+ y^2+z= 1 as a "level surface" for F(x,y,z)= x^2+ y^2+ z. The grad F= 2xi+ 2yj+ k is perpendicular to the surface and (2xi+ 2yj+ k)dydx is the "vector differential of surface area".
The integral of curl v over the surface is the integral of (-i-j-k)*(2xi+2yj+ k)dydx= integral of (-2x-2y-1)dydx. You do not need to find the unit normal. Many text books use "unit normal vector" times dS but dS turns out to include the length of the vector itself- its much simpler just to use the gradient. z=1- x^2- y^2 will be "greater than or equal to 0" as long as x^2+ y^2 is less than or equal to 1: inside the unit circle. You can do the integration in polar coordinates: dA= rdrd&theta; and -2x-2y-1= -2rcos&theta;-2rsin&theta;-1.