Use Green's Theorem to evaluate int_C(e^x+y^2)dx+(e^y+x^2)dy where C is the boundary of the region(traversed counterclockwise) in the first quadrant bounded by y = x^2 and y = 4.

Use Green's Theorem to evaluate ${\int }_{C}\left({e}^{x}+{y}^{2}\right)dx+\left({e}^{y}+{x}^{2}\right)dy$ where C is the boundary of the region(traversed counterclockwise) in the first quadrant bounded by $y={x}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=4$.
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Step 1
The give integral is ${\int }_{C}\left({e}^{x}+{y}^{2}\right)dx+\left({e}^{y}+{x}^{2}\right)dy$ ….(1)
where C is the boundary of the region in first quadrant bounded by $y={x}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=4$.
From greens theorem we know that
${\int }_{C}Pdx+Qdy=\int {\int }_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$ ....(2)
Comparing we get
$P={e}^{x}+{y}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Q={e}^{y}+{x}^{2}$
So,
$\frac{\partial P}{\partial y}=2y\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial Q}{\partial x}=2x$ ...(3)
Step 2
From (1), (2) and (3) we get
${\int }_{C}\left({e}^{x}+{y}^{2}\right)dx+\left({e}^{y}+{x}^{2}\right)dy=\int {\int }_{D}2\left(x-y\right)dxdy$ ...(4)
Now, we will find the region of integration.
$y={x}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=4$
this gives
${x}^{2}=4$
$x=+-2$
Since the region lies on first quadrant so we leave negative sign.
Therefore, x varies from 0 to 2 and y varies from 4 to x2
Now, from (4)
Step 4
${\int }_{C}\left({e}^{x}+{y}^{2}\right)dx+\left({e}^{y}+{x}^{2}\right)dy={\int }_{x=0}^{2}{\int }_{y=4}^{{x}^{2}}2\left(x-y\right)dxdy$
$=2{\int }_{x=0}^{2}{\left(xy-\frac{{y}^{2}}{2}\right)}_{4}^{{x}^{2}}dx$
$=2{\int }_{x=0}^{2}\left(x×{x}^{2}-\frac{{\left({x}^{2}\right)}^{2}}{2}\right)-4x+8\right)dx$
$=2{\int }_{x=0}^{2}\left({x}^{3}-\frac{{x}^{4}}{2}-4x+8\right)dx$