Use Green's Theorem to evaluate the line integral. (2x + ln 2)dy - (8y^2 + sinhx)dx

Question
Use Green's Theorem to evaluate the line integral.
\(\displaystyle{\left({2}{x}+{\ln{{2}}}\right)}{\left.{d}{y}\right.}-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}{\left.{d}{x}\right.}\)

Answers (1)

2020-12-14
Step 1
Let's compare the given integrand with the standard form.
\(\displaystyle{\left({2}{x}+{\ln{{2}}}\right)}{\left.{d}{y}\right.}-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}{\left.{d}{x}\right.}={Q}.{\left.{d}{y}\right.}+{P}.{\left.{d}{x}\right.}\)
Hence, \(\displaystyle{Q}={2}{x}+{\ln{{2}}}\)
and \(\displaystyle{P}=-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}\)
Step 2
Please see the white board.
\(\displaystyle{P}=-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}\)
\(\displaystyle\Rightarrow\frac{{\partial{P}}}{{\partial{y}}}=-{16}{y}\)
\(\displaystyle{Q}={\left({2}{x}+{\ln{{2}}}\right)}\)
\(\displaystyle\Rightarrow\frac{{\partial{Q}}}{{\partial{x}}}={2}\)
Step 3
Limits of integration for x will be from x = 3 to x = 6 (the x corrdinates of the square)
and limits of integration for y will be from y = 1 to y = 4 (the y coordinates of the square)
\(\displaystyle{I}=\oint_{{C}}{\left({P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}\right)}=\int\int_{{R}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{d}{A}\)
\(\displaystyle=\int\int_{{R}}{\left({2}+{16}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{1}}^{{4}}}{\left({2}+{16}{y}\right)}{\left.{d}{y}\right.}{\int_{{3}}^{{6}}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left[{2}{y}+{8}{y}^{{2}}\right]}_{{1}}^{{4}}}\times{{\left[{x}\right]}_{{3}}^{{6}}}\)
\(\displaystyle={126}\times{3}={378}\)
Step 4
Hence, the final answer is:
Given integral = 378
0

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