Question

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

it does seems to be conservative

Answer 5

doesn't

Answer 6

x =/2x

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

ok

Answer 11

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

Answer 12

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.

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

dl = (-i + j) dt

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

ok, thanks