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 2 years 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
 2 years 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

This Question is Closed

modphysnoob
 2 years ago
Best ResponseYou've already chosen the best response.03.51) 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

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1No, what you just wrote for part a is wrong. You are evaluating a line integral.

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1The 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.

modphysnoob
 2 years ago
Best ResponseYou've already chosen the best response.0it does seems to be conservative

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Right, it has nonzero curl. So evaluate the line integral along each of the three edges.

modphysnoob
 2 years ago
Best ResponseYou've already chosen the best response.0ok , 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

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1...where y = 0. Hence the integral = 0

modphysnoob
 2 years ago
Best ResponseYou've already chosen the best response.0so 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

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Do 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.

modphysnoob
 2 years ago
Best ResponseYou've already chosen the best response.0oh, I am doing 3.51, which uses the figure 3.50)b)

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

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

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

modphysnoob
 2 years ago
Best ResponseYou've already chosen the best response.0I took calc 3 long time ago, never thought it would be useful in real life

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1I'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

modphysnoob
 2 years ago
Best ResponseYou've already chosen the best response.0Integrate[(2t)t+(2t)^2+2t^2,{t,0,1}

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