Use Green's Theorem to evaluate the line integral int_C(y+e^x)dx+(6x+cosy)dy where C is triangle with vertices (0,0),(0,2)and(2,2) oriented counterclockwise. a)6 b)10 c)14 d)4 e)8 f)12

Use Green's Theorem to evaluate the line integral int_C(y+e^x)dx+(6x+cosy)dy where C is triangle with vertices (0,0),(0,2)and(2,2) oriented counterclockwise. a)6 b)10 c)14 d)4 e)8 f)12

Question
Use Green's Theorem to evaluate the line integral
\(\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}\)
where C is triangle with vertices (0,0),(0,2)and(2,2) oriented counterclockwise.
a)6
b)10
c)14
d)4
e)8
f)12

Answers (1)

2021-03-13
Step 1
Given \(\displaystyle{P}={y}+{e}^{{x}}{\quad\text{and}\quad}{Q}={6}{x}+{\cos{{y}}}\)
Green's theorem is
\(\displaystyle\int_{{C}}{P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}=\int\int_{{C}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle=\int\int_{{C}}{\left({6}-{1}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={5}\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
Since C is closed triangle with vertices (0,0) , (0,2) & (2,2) which is oriented counter clockwise .
Since \(\displaystyle\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\) is the area of that triangle.
Step 2
Then
\(\displaystyle\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}=\frac{{1}}{{2}}\cdot{2}\cdot{2}={2}\)
Hence
\(\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}={5}\cdot\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
=5*2=10
Therefore (2) option is correct.
0

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