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I am doing practice problem and have no solution to check my work. so I would post my work and please tell me whether or not it is correct

Mathematics
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3.51) a) Since the triangle is in x -y plane , its vector would be in z direction. but E is only in x and y direction , hence it is 0 @JamesJ
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No, what you just wrote for part a is wrong. You are evaluating a line integral.
The first thing I would check is if this E field is conservative, i.e., curl-free. If so, then the line integral over any closed loop must be zero. If E is not conservative, then you will have to calculate the three pieces explicitly.

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Other answers:

it does seems to be conservative
doesn't
x =/2x
Right, it has non-zero curl. So evaluate the line integral along each of the three edges.
ok , so we start with horitozotal line along x axis Integrate[E dl] E= {xy,-x^2-2y^2} dl that part is {1 , 0} Int[ x*y dx ] from 0 to 2
...where y = 0. Hence the integral = 0
ok
so left diagonal would be {1,1}, not sure if I need to normalize it? {xy,-x^2-y^2} doing dot product would result in same in scalar {xy -x^2-y^2} Integreate[xy-x^2-y^2 ] x=0 to 1 , y =0 to 1
Do the vertical piece next. The easiest way to do this and avoid confusion is to explicitly parameterize the curve over which you are integrating. For the vertical part, you are integrating over the portion of the line (1,t), where t = 0 -> 1 Hence the integral over that line is \[ \int_0^1 -(x^2 + 2y^2) \ dt = \int_0^1 -(1 + 2t^2) \ dt \] For the diagonal, use the same idea of parameterization.
oh, I am doing 3.51, which uses the figure 3.50)b)
Ok, in which case a perfectly good parameterization for the line from (2,0) to (1,1) is (2-t,t), t = 0 -> 1
so in that case , you are assigning 2-t to x and t to y right? but what happen to dot product between E and dl
dl = (-i + j) dt
Sounds to me like you should go back to your maths textbook or something and review line integrals
I took calc 3 long time ago, never thought it would be useful in real life
I'll help finish this part and then I'm out of here. E(t) = xy i + -(x^2 + 2y^2) j = (2-t)t i - ( (2-t)^2 + 2t^2 ) j Now calculate E(t). dl and integrate that expression from t = 0 to 1
Integrate[(2-t)t+(2-t)^2+2t^2,{t,0,1}
No, dl = (-i + j) dt hence E(t) . dl = -(2-t)t + - ( (2-t)^2 + 2t^2 )
ok, thanks

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