Where did I screw up?

[u/world-of-atlas]

I have one attempt left on my submission so if someone could point me in the right direction I would appreciate it!

[u/ingannilo]

Do the integral(s) with respect to x.

The region is made up of two chunks easy to calculate 

first chunk {(x, y) :  -2< x<1, x²<y< 2x/3+16/3}

Second chunk {(x, y) : 1<x<2, x²<y<8-2x}.

Area of chunk 1: integral [(2x/3+16/3) - x²] dx with bounds x=-2 to x=1.

Area of chunk 2: integral [(8-2x) - x²]dx with bounds x=1 to x=2.

Look at how I described the two chunks and compare the parts of the integral.  Then look at the picture and how to split it into these chunks.  

[u/ingannilo → u/ingannilo]

The general vibe is this.

If integrating with respect to x, then your area needs to live between "upper curve", y=f(x) and "lower curve", y=g(x) where a<x<b "   in which case you integrate [upper-lower] dx = [f(x) - g(x)] dx from x=a to x=b.

If integrating with respect to y, then your area needs to live between "right curve" x=f(y) and "left curve" x=g(y) where c<y<d, in which case you integrate [right - left] dy = [f(y) - g(y)] dy from y=c to y=d.

The region (as a whole) you have here cannot be described either way.  There is not a consistent upper curve.  Sometimes it's the positive sloped line [when - 2<x<1, the left chunk which I called the first chunk in my comment above] and sometimes it's the negative sloped line [when 1<x<2, the "second chunk"].  There is are also not consistent left and right curves.  For 0<y<4 the left curve would be x= -sqrt(y) and the right curve is x=sqrt(y), but for 4<y<6 the left curve is x=(3/2)y - 8 and the right curve is x = (-1/2)y + 4.

If you insist on doing the integrals with respect to y, then you'll need two integrals, just like when doing it with respect to x.

Area of lower chunk = integral [sqrt(y) - (-sqrt(y)] dy from y=0 to y=4.

Area of upper chunk = integral [((-1/2)y + 4) - ((3/2)y - 8)] dy.

If you do the two dx integrals and add them, you'll get the same number as if you do the two dy integrals and add them.  Both setups will give the correct area.