Question

Evaluate the line integral oint_C xy dx + x^2 dy, where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3), and (0,3). You can evaluate directly or use Green's theorem. Write the integral(s), but do not evaluate.

Evaluate the line integral \(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}\), where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3), and (0,3). You can evaluate directly or use Green's theorem.
Write the integral(s), but do not evaluate.

Answers (1)

2020-12-03
Step 1
Given,
\(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}\), Where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3),(0,3)
Step 2
Green's theorem:
Let C be the simply connected region with piecewise smooth boundary C oriented counterclockwise,
If M and N are the first partial derivative in an open region containing C, then
\(\displaystyle\oint_{{C}}{M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}=\int_{{R}}\int\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}{d}{A}\)
Consider, \(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}\)
M=xy
\(\displaystyle{N}={x}^{{2}}\)
\(\displaystyle\frac{{\partial{M}}}{{\partial{y}}}={x}\)
\(\displaystyle\frac{{\partial{N}}}{{\partial{x}}}={2}{x}\)
Limits of x are \(\displaystyle{0}\le{x}\le{2}\)
Limits of y are \(\displaystyle{0}\le{y}\le{3}\)
Now the integral becomes,
\(\displaystyle{\int_{{0}}^{{3}}}{\int_{{0}}^{{2}}}{\left({2}{x}-{x}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{3}}}{\int_{{0}}^{{2}}}{\left({x}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{{\left(\frac{{x}^{{2}}}{{2}}\right)}_{{0}}^{{2}}}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{\left(\frac{{2}^{{2}}}{{2}}-\frac{{0}}{{2}}\right)}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}\frac{{1}}{{4}}{\left.{d}{y}\right.}\)
\(\displaystyle=\frac{{1}}{{4}}{\int_{{0}}^{{3}}}{\left.{d}{y}\right.}\)
\(\displaystyle=\frac{{1}}{{4}}{{\left({y}\right)}_{{0}}^{{3}}}\)
\(\displaystyle=\frac{{1}}{{4}}{\left({3}-{0}\right)}\)
\(\displaystyle=\frac{{3}}{{4}}\)
Therefore, \(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}=\frac{{3}}{{4}}\)
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