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## Homework Statement

Assume the vector function A = ax(3x[tex]^{2}[/tex]2y[tex]^{2}[/tex])-ax(x[tex]^{3}[/tex]y[tex]^{2}[/tex])

a) Find [tex]\oint[/tex]A[tex]\cdot[/tex]dl around the triangular contour shown in Fig. 2-36 [it is a triangle with base and height of one on the x and y axis. the curl travels so that the normal vector is in the -z direction]

b) Evaluate [tex]\int[/tex]([tex]\nabla[/tex][tex]\times[/tex]A)[tex]\cdot[/tex]ds over the triangular area.

## Homework Equations

The equation in part a) should be equivalent to that of b) as per Stoke's theorem

## The Attempt at a Solution

Part a) is where I'm having trouble. Here is my work for that and I am 100% sure it is incorrect. I have no idea how to set it up, that's my problem. The execution is no big deal, but the set-up part is. Any guidance is appreciated.

A[tex]\cdot[/tex]dl = [ax(3x[tex]^{2}[/tex]2y[tex]^{2}[/tex])-ax(x[tex]^{3}[/tex]y[tex]^{2}[/tex]) [tex]\cdot[/tex] [ax(dx) + ay(dy)]

= 3x[tex]^{2}[/tex]y[tex]^{2}[/tex]dx - x[tex]^{3}[/tex]y[tex]^{2}[/tex]dy

[tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex]3x[tex]^{2}[/tex]y[tex]^{2}[/tex]dx - [tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex]x[tex]^{3}[/tex]y[tex]^{2}[/tex]dy

I then said the area of the triangle is equal to 1/2 (since .5*1*1 = .5) From there I said x = 1/y and y =1/x so I could plug into each equation to get:

[tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex]3x[tex]^{2}[/tex](1/y)[tex]^{2}[/tex]dx - [tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex](1/x)[tex]^{3}[/tex]y[tex]^{2}[/tex]dy

= 3 - ln(2) =2.3068

PART B:

I'm fairly confident in my execution of this, but it would be worth verifying.

[tex]\nabla[/tex][tex]\times[/tex]A = ax([tex]\partial[/tex]/[tex]\partial[/tex]y * Az - [tex]\partial[/tex]/[tex]\partial[/tex]z * Ay) + ay([tex]\partial[/tex]/[tex]\partial[/tex]z * Ax - [tex]\partial[/tex]/[tex]\partial[/tex]x * Az) + az([tex]\partial[/tex]/[tex]\partial[/tex]x * Ay - [tex]\partial[/tex]/[tex]\partial[/tex]y * Ax)

= ax(0-0) + ay(0-0) + az(-6x[tex]^{4}[/tex]y[tex]^{4}[/tex] + 6x[tex]^{5}[/tex]y[tex]^{3}[/tex])

ds = ds[tex]_{z}[/tex] = dxdy

[tex]\int[/tex]([tex]\nabla[/tex][tex]\times[/tex]A)[tex]\cdot[/tex]ds = [tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex][tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex](-6x[tex]^{4}[/tex]y[tex]^{4}[/tex] + 6x[tex]^{5}[/tex]y[tex]^{3}[/tex])dxdy = 5.61

EDIT- for some reason the tex stuff is being really obnoxious for me. the 12 that your seeing is the boundry of the integral (going from 1 to 2) and if you look carefully, there is a dot in between A and dl and others like that. I apologize for it looking crappy!

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