# Green's Theorem

Let $C$ be the closed,piecewise curve figured by traveling in straight lines between the points $(-2,1),(-2,-3),(1,-1),(1,5)$ and back to $(-2,1)$, in that order. Use Green's Theorem to evaluate the integral:

$$\int_C (2 x y)dx + (x y^2)dy$$

So far I have used Green's Theorem and calculated the vector form of Green's Theorem. Since the region enclosed by C is y-simple, we can set up the double integral to first integrate in terms of $y(x)$ then $x$. However, I am not getting the same answer as the book.

• I was just wondering if how I am approaching the question wrong or if its simply a calculation erorr – tamefoxes Mar 23 '13 at 7:36
• What answer did you get? – Jesse Madnick Mar 23 '13 at 8:26

The region in question is a trapezoid, bounded by the lines $x=-2$, $x=1$, $y=(4/3) x+(11/3)$, and $y=(2/3)x-(5/3)$. Use Green's theorem to convert the line integral to an integral over the area of the region:

$$\oint_{\partial D} (P\, dx+Q\, dy) = \iint_D dx\,dy \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$$

To set up the area integral, take the derivatives and use those bounds:

$$\int_{-2}^1 dx \: \int_{(2/3)x-(5/3)}^{(4/3) x+(11/3)} dy \: (y^2-2 x)$$

You should be able to work this out.

• For the line y=2x+3 I got something slightly different, however these are the steps that I took to get an answer. Thanks for the clarification! – tamefoxes Mar 23 '13 at 19:17
• Did you get $(4/3) x+(11/3)$? – Ron Gordon Mar 23 '13 at 19:21
• Yes that was my upper bound for y(x) – tamefoxes Mar 23 '13 at 21:32
• Very sorry for the error then. – Ron Gordon Mar 23 '13 at 22:09
• It's fine, it happens once in a while! – tamefoxes Mar 24 '13 at 5:36