Find a vector field F whose curl is G and use Stokes’ theorem

[Colt87]

Find a vector field F whose curl is G and use Stokes’ theorem

Consider the vector fieldG(x, y, z) = xi −y/ 2j + ( 2xy −z/ 2 ) k,and a wire frame described byC = { (x, y, z)|x² + y² = 2 and z = 1}

I found the parameters
r(s,t) = (Sqrt 2 cos(s), Sqrt 2 sin(s), t)

from 0<=s<=2pi and

0<=t<=1 (not sure about this bound)

I started with the curl = del X F

but how do i find F(xyz) to get (del X F = G) from that?

 

[HallsofIvy]

Frankly I don't know what you mean by "a wire frame described by C = \(\displaystyle \{ (x, y, z)|x^2 + y^2 = 2 and z = 1\}\). That is a circle, in the z= 1 plane, with center at (0, 0, 1) and radius \(\displaystyle \sqrt{2}\). And, that does not seem to have anything to do with your question. You are given the vector field \(\displaystyle G= x\vec{i}+ \frac{y}{2}\vec{k}+ \left(2xy- \frac{z}{2}\right)\vec{k}\). You want to find F such that \(\displaystyle \nabla\times F= G\).

Okay, Let \(\displaystyle F= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}\). Then \(\displaystyle \nabla\times F\) is
\(\displaystyle \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f & g & h \end{array}\right|= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}\).

So we must have \(\displaystyle \frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}= x\), \(\displaystyle \frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}= \frac{y}{x}\), and \(\displaystyle \frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}= 2xy- \frac{z}{2}\).

Now, what do you want to do with this? You say "use Stokes' theorem". Use Stokes' theorem to do what?

 

[Colt87]

Frankly I don't know what you mean by "a wire frame described by C = \(\displaystyle \{ (x, y, z)|x^2 + y^2 = 2 and z = 1\}\). That is a circle, in the z= 1 plane, with center at (0, 0, 1) and radius \(\displaystyle \sqrt{2}\). And, that does not seem to have anything to do with your question. You are given the vector field \(\displaystyle G= x\vec{i}+ \frac{y}{2}\vec{k}+ \left(2xy- \frac{z}{2}\right)\vec{k}\). You want to find F such that \(\displaystyle \nabla\times F= G\).

Okay, Let \(\displaystyle F= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}\). Then \(\displaystyle \nabla\times F\) is
\(\displaystyle \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f & g & h \end{array}\right|= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}\).

So we must have \(\displaystyle \frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}= x\), \(\displaystyle \frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}= \frac{y}{x}\), and \(\displaystyle \frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}= 2xy- \frac{z}{2}\).

Now, what do you want to do with this? You say "use Stokes' theorem". Use Stokes' theorem to do what?

Right sorry so i'm pretty sure it's a cylinder. I found F(xyz) = 1/2zyi + x²yj +xyk which G is it's curl.

This is the full question:

Consider the vector field:

G(x, y, z) = xi −y/2j + (2xy −z/2)k,

and a wire frame described by

C = {(x, y, z)|x² + y² = sqrt2 and z = 1,

with counter clock wise orientation when viewed from above. Suppose thatS is any smooth surface with boundary C. Compute the flux of G throughS. (Hint: find a vector field F whose curl is G and use Stokes’ theorem.You will need to figure out a parameterization of C to do the resulting lineintegral).