I am having trouble with the follow problem about Stoke's theorem:
Assume the vector field $A= <3x^2y^3,−x^3y^2>$ Verify Stoke's theorem for the closed path around the triangle with vertices (1,1) (2,1) (2,2) by:
a.) Finding $∮A·dl$
b.) Evaluating $∫∫(∇×A)·dS$
I have attempted to solve using both methods but have ended up with differing answers. for part A, I parametrized the path between $<t,1>,<2,t>, <t,t>$ for the triangle, and ended up with three seperate integrals, $∫ 3t^2dt + ∫ 3t^5dt+ ∫ -8t^2dt$, and evaluated them from (2,1) and got the answer as $-8.33$
For part B, I found the curl, $(∇×A)$ to be $-12x^2y^2$ and evaluated the integral $∫∫(-12x^2y^2)dxdy$ from (2,1), (2,1), but got the answer as $-65.3$
Where am I going wrong here?
Stokes' theorem for a region $R$ is equivalent to Green's theorem in the plane, so that we need to show
$$\oint_C(Mdx+Ndy)=\iint_R\Big(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\Big)dxdy$$
where $R$ is the region bounded by the triangle with vertices $(1,1)$, $(2,1)$ and $(2,2)$; $C$ is the boundary of $R$. Here, $M=3x^2y^3$ and $N=−x^3y^2$.
Along the line from $(1,1)$ to $(2,1)$, $y=1$ ($dy=0$) and $x$ varies from $1$ to $2$ so that
$$\int_{C_1}(Mdx+Ndy)=\int_1^23x^2dy=7.$$
Along the line from $(2,1)$ to $(2,2)$, $x=2$ and $y$ varies from $1$ to $2$:
$$\int_{C_2}(Mdx+Ndy)=-8\int_1^2y^2dy=-\frac{56}{3}.$$
Along the line from $(2,2)$ to $(1,1)$, $x=y$ so that
$$\int_{C_3}(Mdx+Ndy)=\int_2^12x^5dx=-\frac{63}{3}.$$
Therefore,
$$\oint_C(Mdx+Ndy)=-\frac{98}{3}.$$
Also, the region $R$ bounded by the triangle is $\frac{1}{2}$ of the region bounded by the square with adjacent sides $C_1$ and $C_2$ so that
$$\iint_R\Big(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\Big)dxdy=\frac{1}{2}\int_1^2\int_1^2-12x^2y^2dxdy=-\frac{98}{3}.$$