Question

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 \(\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}\) where C is the boundary of the region(traversed counterclockwise) in the first quadrant bounded by \(\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}\).

Answers (1)

2021-02-26

Step 1
The give integral is \(\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}\) ….(1)
where C is the boundary of the region in first quadrant bounded by \(\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}\).
From greens theorem we know that
\(\displaystyle\int_{{C}}{P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}=\int\int_{{D}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\) ....(2)
Comparing we get
\(\displaystyle{P}={e}^{{x}}+{y}^{{2}}{\quad\text{and}\quad}{Q}={e}^{{y}}+{x}^{{2}}\)
So,
\(\displaystyle\frac{{\partial{P}}}{{\partial{y}}}={2}{y}{\quad\text{and}\quad}\frac{{\partial{Q}}}{{\partial{x}}}={2}{x}\) ...(3)
Step 2
From (1), (2) and (3) we get
\(\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}=\int\int_{{D}}{2}{\left({x}-{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\) ...(4)
Now, we will find the region of integration.
\(\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}\)
this gives
\(\displaystyle{x}^{{2}}={4}\)
\(x=+-2\)
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
\(\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}={\int_{{{x}={0}}}^{{2}}}{\int_{{{y}={4}}}^{{{x}^{{2}}}}}{2}{\left({x}-{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={2}{\int_{{{x}={0}}}^{{2}}}{{\left({x}{y}-\frac{{y}^{{2}}}{{2}}\right)}_{{4}}^{{{x}^{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={2}{\int_{{{x}={0}}}^{{2}}}{\left({x}\times{x}^{{2}}-\frac{{{\left({x}^{{2}}\right)}^{{2}}}}{{2}}\right)}-{4}{x}+{8}{)}{\left.{d}{x}\right.}\)
\(\displaystyle={2}{\int_{{{x}={0}}}^{{2}}}{\left({x}^{{3}}-\frac{{x}^{{4}}}{{2}}-{4}{x}+{8}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={2}{{\left(\frac{{x}^{{4}}}{{4}}-\frac{{x}^{{5}}}{{10}}-{2}{x}^{{2}}+{8}{x}\right)}_{{0}}^{{2}}}\)
\(\displaystyle={2}{\left(\frac{{{4}-{16}}}{{{5}-{8}+{16}}}\right)}\)
\(\displaystyle={2}{\left({12}-\frac{{16}}{{5}}\right)}\)
\(\displaystyle={2}{\left(\frac{{{60}-{16}}}{{5}}\right)}\)
\(\displaystyle={2}{\left(\frac{{44}}{{5}}\right)}\)
\(\displaystyle=\frac{{88}}{{5}}\)

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