What is the proper way to set up this integral to find the area bounded by the curves? 
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle. Then find the area.
$$y = 2x^2, y = 8x^2, 3x + y = 5, x ≥ 0$$ 

Here is my drawing: http://www.webassign.net/waplots/d/a/986f170c68ee080b539049f410cbba.gif 
If I'm integrating with in terms of x, I know I'm going to need more than one integral, but I don't know how to set it up. Any help is appreciated

You can use double integral, 
the intersection points: 
between $y=8x^2$ and $y=5-3x$ are at $x=-1$ and $x=5/8$
between $y=2x^2$ and $y=5-3x$ are at $x=-5/2$ and $x=1$
let's assume that we want $\text{d$y$d$x$}$ 
for $I_1$: when $y$ goes from $y=2x^2$ to $y=8x^2$ then $x$ goes from $0$ to $5/8$
for $I_2$: when $y$ goes from $y=2x^2$ to $y=5-3x$ then $x$ goes from $5/8$ to $1$
$$\underbrace{\int_{x=0}^{x=5/8}\int_{y=2x^2}^{y=8x^2}1\text{d$y$d$x$}}_{I_1}+\underbrace{\int_{x=5/8}^{x=1}\int_{y=2x^2}^{y=5-3x}1\text{d$y$d$x$}}_{I_2}$$
$$=\int_{0}^{5/8} \left(8x^2 - 2x^2\right)\text{d}x + \int_{5/8}^{1} \left(-3x + 5 - 2x^2\right)\text{d}x = \frac{125}{256}+\frac{117}{256}$$
