Find $\displaystyle [integral](x^2)*(y^2)dx + 4*x*y^3dy$ where C is the boundary of the triangle with the corners (0,0), (1,3), and (0,3). Mainly how do I find the limits of integration?
The problem is a path integral around a closed path. It would not be difficult to integrate but, since the path is not smooth would require three different integrals. Green's theorem allows that to be done as a single integral over the region enclosed by the path.
Sgraveyard, as dwsmith suggested, first draw the picture. It is, of course, a triangle having (0, 0), (1, 3), and (0, 3) as vertices.
The limits of integration depend on the order of integration. If you decide to integrate with respect to y first, then x, you need to look at the smallest and largest possible values for x. They are, of course, 0 and 1. Now, draw a vertical line representing "x= constant". For each x, that line crosses the triangle at y= 3 above and on the line from (0, 0) to (1, 3) below. Write the equation for that line and solve for y. That will be the lower bound on the y-integral:
$\displaystyle \int_{x= 0}^1 \int_{y= ax+b}^3 dydx$
(You need to put in the "a" and "b".)
Or you could decide to integrate with respect to x first and then y, look at the smallest and largest values of y- they are clearly 0 and 3. Draw a horizontal line, representing y= constant. That will cross the triangle at x= 0 and on the line from (0,0) to (1, 3) on the right. Write the equation for that line and solve for x. That will be the upper bound on the x-integral:
$\displaystyle \int_{y= 0}^3\int_{x= 0}^{my+ c} dxdy$