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

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- anonymous

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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- JamesJ

No, what you just wrote for part a is wrong. You are evaluating a line integral.

- JamesJ

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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- anonymous

it does seems to be conservative

- anonymous

doesn't

- anonymous

x =/2x

- JamesJ

Right, it has non-zero curl. So evaluate the line integral along each of the three edges.

- anonymous

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

- JamesJ

...where y = 0. Hence the integral = 0

- anonymous

ok

- anonymous

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

- JamesJ

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.

- anonymous

oh, I am doing 3.51, which uses the figure 3.50)b)

- JamesJ

Ok, in which case a perfectly good parameterization for the line from (2,0) to (1,1) is
(2-t,t), t = 0 -> 1

- anonymous

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

- JamesJ

dl = (-i + j) dt

- JamesJ

Sounds to me like you should go back to your maths textbook or something and review line integrals

- anonymous

I took calc 3 long time ago, never thought it would be useful in real life

- JamesJ

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

- anonymous

Integrate[(2-t)t+(2-t)^2+2t^2,{t,0,1}

- JamesJ

No, dl = (-i + j) dt hence
E(t) . dl = -(2-t)t + - ( (2-t)^2 + 2t^2 )

- anonymous

ok, thanks

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