Use Green's Theorem to Find the Area Use Green's Theorem to find the area enclosed by: 

$$y=9-x^{2},y=8x, y=\frac{2}{5}x$$

(The area in Quadrant 1)
In class we only did examples of this type of problem that were very simple (eg. area under $\ x^{2}\ $ from 0 to 2), which made setting up the equation for area using Green's Theorem simple. But since this problem has multiple lines/curves to consider, I am not sure how to set up the problem once I am done paramaterizing. 
 A: HINT:
For a region $Ω$ and a 1-form $ω=P(x,y)dx+Q(x,y)dy$ notice that Green's theorem says $\int_{\partial Ω}ω=\int_Ωdω$ (where $dω=(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy$. If you don't know about differential forms just see that what I have writen makes sense symbolicaly.)
Now notice that if $ω=-ydx+xdy$ then $dω=2dxdy$ therefore integrating $ω$ over the boundary of a region (with the correct orientation) will give you twice the area of that region.
A: This really isn’t any different from the simpler examples that you’ve seen, or, for that matter, from the way you’d approach this in first-year Calculus: find the intersections of the curves and integrate piecewise. Remember that even in the example you cite of finding the area under $x^2$, the region is bounded by three curves: that one and the lines $y=0$ and $x=2$. The intersections of these curves are only slightly easier to find than the ones for the region in your question.  

The region is bounded below by the line $y=\frac25x$ and above by $y=8x$ and $y=9-x^2$. One intersection point is obviously the origin, and the other two are easily found by substitution into the quadratic. You’ll then need to integrate along the lower line segment from the origin to the first intersection point, along the parabola to the second, and then back to the origin along the other line segment. Take care to orient the curves correctly.
