Surface integral of curl A over paraboloid z=1-x²-y²

[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!

 

[HallsofIvy]

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.

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 textbooks 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= rdrdθ and -2x-2y-1= -2rcosθ-2rsinθ-1.

 

[M98Ranger]

Hi there,

First of all, great job setting up the problem correctly! You are on the right track and your understanding of Stoke's Theorem is correct. Now let's go through the steps to find the surface integral of (grad x A)*dA.

1. Finding the curl of A: You have correctly found the curl of A to be -i+j-k using the determinant method.

2. Finding the unit normal vector: To find the unit normal vector, we need to take the gradient of the surface S. However, it seems like you have made a small mistake in rearranging the surface equation. It should be x^2+y^2+z=1 instead of x^2+y^2+z=1-x^2-y^2. This will give us the gradient of the surface as 2xi+2yj+k. Now, to find the unit normal vector, we divide this by its magnitude which is √(4x^2+4y^2+1). However, this is not the final answer. We need to divide each component by this magnitude to get the unit normal vector as (2x/√(4x^2+4y^2+1))i + (2y/√(4x^2+4y^2+1))j + (1/√(4x^2+4y^2+1))k.

3. Finding dS: As you correctly mentioned, dS is given by (dx*dy)/|n*k|. However, we need to make sure to use the unit normal vector we just found in step 2. So, dS will be (dx*dy)/√(4x^2+4y^2+1).

4. Setting up the double integral: Now that we have all the components, we can set up the double integral as ∫∫(grad x A)*n*dS. In polar coordinates, this will be ∫∫(grad x A)*n*r dr dθ, where the limits of integration will be 0 to 1 for both r and θ.

5. Evaluating the integral: Finally, we can plug in the values for (grad x A) and n, and evaluate the integral. This will give us the surface integral of (grad x A)*dA.

I hope this helps you in solving the problem. Don