Find integral area with respect to x The question is finding the area enclosed by the curves x=$y^2$ and x+2y=8 using both x and y integrals
Graph for reference Purple is x+2y=8, red is x=$y^2$ 
First I found the limits by letting x=8-2y. This gave the equation $y^2$+2y-8 which gave y=-4 and y=2. 
Putting back into equation gives x=4 and x=16
I calculated with y integral being
$$\int_4^2 (8-2y)-y^2 \,$$  ( lower integral is -4, I can't seem to express)
Now I would like to ask how would you calculate using the x integral
I had a think about this and got 

but the answers x and y integrals are different so I think theres a mistake somewhere but I don't know what I did wrong.

The first should be $$\int\limits_{-4}^2(8-2y-y^2)dy.$$
The second should be $$\int\limits_0^4\left(\sqrt{4-\frac{x}{2}}-\left(-\sqrt{4-\frac{x}{2}}\right)\right)dx+\int\limits_4^{16}\left(4-\frac{x}{2}-\left(\sqrt{4-\frac{x}{2}}\right)\right)dx$$
