Solve Stoke's Theorem with Contour C and Vector Field F for Calculus Homework

[null void]

Homework Statement

\oint_C{(x^2 + 2y + sin x^2)dx + (x + y + cos y^2)dy}

the contour C formed by 3 curves:
C(x,y) = \begin{cases}x=0, \quad from (0,0) to (0,5)\\y = 5 - x^2,\quad from(0,5) to (2,1) \\ 4y = x^2, \quad from(2,1) to (0,0)\end{cases}

and the Stoke Theorem:
\oint_C \vec F \, d\vec r = \iint_S \, curl \, \vec F \, d\vec S \\= \iint_S \, curl \, \vec F \cdot \vec n \, dS

The Attempt at a Solution

So in this problem, can I say the vector field, F as follow?
\vec F = <x^2 + 2y + sin x^2 ,\,\, x + y + cos y^2, \,\, 0>
then
\oint_C \vec F \, d\vec r = \iint_S \, curl \, \vec F \, d\vec S \\ = \int_0^5 \int_\frac{x^2}{4}^{5-x^2} -1\, dy\, dx \\ = 27.0833

but probably there is something wrong in there, because the answer is 20/3

 

[RUber]

Your x only ranges from 0 to 2, not 0 to 5.

 

[null void]

Oh yeah, what a silly mistake, thanks.

 

[Ray Vickson]

Homework Statement

\oint_C{(x^2 + 2y + sin x^2)dx + (x + y + cos y^2)dy}

the contour C formed by 3 curves:
C(x,y) = \begin{cases}x=0, \quad from (0,0) to (0,5)\\y = 5 - x^2,\quad from(0,5) to (2,1) \\ 4y = x^2, \quad from(2,1) to (0,0)\end{cases}

and the Stoke Theorem:
\oint_C \vec F \, d\vec r = \iint_S \, curl \, \vec F \, d\vec S \\= \iint_S \, curl \, \vec F \cdot \vec n \, dS

The Attempt at a Solution

So in this problem, can I say the vector field, F as follow?
\vec F = <x^2 + 2y + sin x^2 ,\,\, x + y + cos y^2, \,\, 0>
then
\oint_C \vec F \, d\vec r = \iint_S \, curl \, \vec F \, d\vec S \\ = \int_0^5 \int_\frac{x^2}{4}^{5-x^2} -1\, dy\, dx \\ = 27.0833

but probably there is something wrong in there, because the answer is 20/3

Your notation is ambiguous and confusing: does $sin x^2$ mean $\sin(x^2)$ or $\sin^2 x = (\sin x)^2$?

 

[null void]

Your notation is ambiguous and confusing: does $sin x^2$ mean $\sin(x^2)$ or $\sin^2 x = (\sin x)^2$?

it is sin(x^2), i will make it clear next time