Is this a good case for horizontal integration? Find enclosed area btw 2 equations. Why or why not? When doing horizontal integration... why do we write the function as a function of y with the x by itself? What is the high level point of that?
So here is my work. Is this right? I am asked to find the area enclosed between these equations:
$$ 4x + y^2 = 12, x = y$$
First thing's first is to graph it and find the points of intersection:

Point of intersection:
$$ \frac{12 - y^2}{4} = y$$
$$ 12 - y^2 = 4y$$
$$y^2 + 4y -12 = 0$$
$$(y - 2)(y+6) = 0$$
$$y = {2,-6}$$
So the points are $(2,2) , (-6, -6)$. 
So the integral...why do we write it as a function of y? Can't we write it as a function of x still?:
$$ \int_{-6}^2 - y \, dy$$
$$= \int_{-6}^2 3 \cdot \frac{-1}{4} y^2 \, dy$$
$$= [ 3y - \frac{1}{4} \cdot \frac{y^3}{3} - \frac{y^2}{2} ]_{-6}^2$$
$$ 6 - \frac{2}{3} - 2 - (-18 + 18 -18) = 21 \frac{1}{3}$$
Does this look right? More importantly, why do we use the equation where y is the variable?
 A: You could write it as a function of whatever you  want; as long as you do the integral correctly, either will give you the area between the two curves.  However, choosing one variable (in this case, $y$) to set up the integral for is sometimes easier than the other. 
Here, for instance, writing this as a function $f(y)$ and integrating with respect to $y$ lets us write this area as a single integral $\int_{-6}^2 f(y)\, dy$. Here your $f(y)$ should be $\frac{12 - y^2}{4} - y$.
However, if we tried to integrate $dx$, we would run into two inconveniences. First, setting it up as a function $g(x)$ of $x$, we would have $y = \pm\sqrt{\text{something}}$, and we would have to make a choice of  sign here. Secondly, when trying to write this integral, we have to integrate the top function minus the bottom function right? But these change; from $x = -6$ to $x =  2$, the top function is $y = x$, and the bottom is the $4x + y^2  = 12$ function.  However, at $x = 2$, we switch; the area is now enveloped on top by the positive square root $y = \sqrt{12 - 4x}$, and on the bottom by the negative square root. 
Thus,  in trying to do the $dx$ integral, we would a) have to work with more annoying functions, and b)  have to split our integral up  into two integrals. It's easier to integrate $dy$!
