Verifying Stokes' theorem in computing line integral

[Win_odd Dhamnekar]

Verify that Stoke’s theorem is true for vector field [math] F = \langle y,x,-z \rangle [/math] and surface S, where S is the upwardly oriented portion of the graph of [math] f(x,y) = x^2y [/math] over a triangle in the x-y plane (0,0), (2,0) and (0,2).

Answer:-

As a surface integral you have [math]f(x,y) = x^2y, curlF =\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & x & -z \end{vmatrix}= [0,0,0] [/math]
[math]\iint\limits_S curl F \cdot dS = \iint\limits_D curl F(x,y,z) \cdot (f_x\times f_y ) dA[/math]
[math] \iint\limits_S curl F \cdot dS = \iint\limits_D [0,0,0] \cdot [-2xy,-x^2, 1] dA =0[/math]
As a line integral, you can parameterize C by [math]r (x,y,z)= (x,y, x^2y), 0\leq x \leq 2, 0\leq y \leq 2[/math]
[math]\int_C F\cdot dr = \int_0^2 \int_0^2 [y + x -(x^2y)(2xy +x^2)] dydx =-\frac{392}{15} [/math]
Where I am wrong? Would any member of this forum tell me in the reply to this thread correctly?

 

[Win_odd Dhamnekar]

Verify that Stoke’s theorem is true for vector field [math] F = \langle y,x,-z \rangle [/math] and surface S, where S is the upwardly oriented portion of the graph of [math] f(x,y) = x^2y [/math] over a triangle in the x-y plane (0,0), (2,0) and (0,2).

Answer:-

As a surface integral you have [math]f(x,y) = x^2y, curlF =\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & x & -z \end{vmatrix}= [0,0,0] [/math]
[math]\iint\limits_S curl F \cdot dS = \iint\limits_D curl F(x,y,z) \cdot (f_x\times f_y ) dA[/math]
[math] \iint\limits_S curl F \cdot dS = \iint\limits_D [0,0,0] \cdot [-2xy,-x^2, 1] dA =0[/math]
As a line integral, you can parameterize C by [math]r (x,y,z)= (x,y, x^2y), 0\leq x \leq 2, 0\leq y \leq 2[/math]
[math]\int_C F\cdot dr = \int_0^2 \int_0^2 [y + x -(x^2y)(2xy +x^2)] dydx =-\frac{392}{15} [/math]
Where I am wrong? Would any member of this forum tell me in the reply to this thread correctly?

I want to make amendment in my answer. As a line integral, you can parameterize C by r(x,y,z) = [x ,-y ,0] [math] \iint_C F\cdot dr = \int_0^2 \int_0^2 [y - x]dydx = 0[/math] . Now, the answer is correct. isn't it?

 

[blamocur]

• While you answer for the curl(F) looks right I don't know if you teacher expects a more detailed explanation.
• As for your parametrization of C, it does not look right to me. Nor can I figure out why you are using double integrals along a curve. Hint: C is piecewise smooth, so you might find it easier to parametrize each piece separately and break the integral along C into separate integrals along each parametrized piece.

 

[Win_odd Dhamnekar]

I computed the double integral as follows:

[math] \displaystyle\int\limits_C F \cdot dr = \displaystyle\int_0^2\displaystyle\int_2^0 [y, x , -z] \cdot [1,-1,0] dt[/math][math]\displaystyle\int\limits_C F \cdot dr =\displaystyle\int_0^2\displaystyle\int_2^0 [y-x] dt[/math][math] \displaystyle\int\limits_C F \cdot dr = 0 [/math]
But you are saying this is wrong. and your hint is correct. Isn't it?

 

[blamocur]

But you are saying this is wrong. and your hint is correct. Isn't it?

I'll reformulate:

1. Why curl(F) = 0?
2. How do you parametrize the curve?

And yes, I think my hint is correct, and, I hope, useful. Otherwise I wouldn't post it.

As for you latest post (#4): sorry, but I simply don't understand what you mean there.

 

[Dr.Peterson]

I computed the double integral as follows:

[math] \displaystyle\int\limits_C F \cdot dr = \displaystyle\int_0^2\displaystyle\int_2^0 [y, x , -z] \cdot [1,-1,0] dt[/math][math]\displaystyle\int\limits_C F \cdot dr =\displaystyle\int_0^2\displaystyle\int_2^0 [y-x] dt[/math][math] \displaystyle\int\limits_C F \cdot dr = 0 [/math]

What does it mean to have two integral signs, but only one differential at the end? What variable do the limits of integration apply to? This really makes no sense at all.

You need to actually express the curve in terms of parameter t. And dr will be different for each part of that boundary "curve", which is a triangle.