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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
 one year ago
 one year ago
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
 one year ago
 one year ago

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modphysnoobBest ResponseYou've already chosen the best response.0
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
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
No, what you just wrote for part a is wrong. You are evaluating a line integral.
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
The first thing I would check is if this E field is conservative, i.e., curlfree. 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.
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
it does seems to be conservative
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
Right, it has nonzero curl. So evaluate the line integral along each of the three edges.
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
ok , so we start with horitozotal line along x axis Integrate[E dl] E= {xy,x^22y^2} dl that part is {1 , 0} Int[ x*y dx ] from 0 to 2
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
...where y = 0. Hence the integral = 0
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
so left diagonal would be {1,1}, not sure if I need to normalize it? {xy,x^2y^2} doing dot product would result in same in scalar {xy x^2y^2} Integreate[xyx^2y^2 ] x=0 to 1 , y =0 to 1
 one year ago

JamesJBest ResponseYou've already chosen the best response.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.
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
oh, I am doing 3.51, which uses the figure 3.50)b)
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
Ok, in which case a perfectly good parameterization for the line from (2,0) to (1,1) is (2t,t), t = 0 > 1
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
so in that case , you are assigning 2t to x and t to y right? but what happen to dot product between E and dl
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
Sounds to me like you should go back to your maths textbook or something and review line integrals
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
I took calc 3 long time ago, never thought it would be useful in real life
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
I'll help finish this part and then I'm out of here. E(t) = xy i + (x^2 + 2y^2) j = (2t)t i  ( (2t)^2 + 2t^2 ) j Now calculate E(t). dl and integrate that expression from t = 0 to 1
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
Integrate[(2t)t+(2t)^2+2t^2,{t,0,1}
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
No, dl = (i + j) dt hence E(t) . dl = (2t)t +  ( (2t)^2 + 2t^2 )
 one year ago
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