0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I was just wondering if how I am approaching the question wrong or if its simply a calculation erorr $\endgroup$ – tamefoxes Mar 23 '13 at 7:36
  • $\begingroup$ What answer did you get? $\endgroup$ – Jesse Madnick Mar 23 '13 at 8:26
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – tamefoxes Mar 23 '13 at 19:17
  • $\begingroup$ Did you get $(4/3) x+(11/3)$? $\endgroup$ – Ron Gordon Mar 23 '13 at 19:21
  • $\begingroup$ Yes that was my upper bound for y(x) $\endgroup$ – tamefoxes Mar 23 '13 at 21:32
  • $\begingroup$ Very sorry for the error then. $\endgroup$ – Ron Gordon Mar 23 '13 at 22:09
  • $\begingroup$ It's fine, it happens once in a while! $\endgroup$ – tamefoxes Mar 24 '13 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.