I'm asked to find $\underset{U}{\int}(x+y)^2\, dA$ where U is a region bounded by the lines
x = -1, x = 1, y = -1
... and by the curves
x=$y^2$ , y=1+$x^2$
Plot: http://d.pr/WYSg
I started out by finding the volume above the box [-1,1] x [-1,2]
$$\underset{-1}{\overset{2}{\int}}\underset{-1}{\overset{1}{\int}}(x+y)^{2}dx\,dy=8$$
And then subtracted the volume above the "upper parabola" y=1+$x^2$: $$\underset{-1}{\overset{1}{\int}}\underset{1+x^{2}}{\overset{2}{\int}}(x+y)^{2}dy\, dx=-\frac{132}{35}$$
As well as the "lower parabola" x=$y^2$: $$\underset{-1}{\overset{1}{\int}}\underset{y^{2}}{\overset{1}{\int}}(x+y)^{2}dx\, dy=-\frac{88}{105}$$
So I get $$8-(-\frac{132}{35})-(-\frac{88}{105})=\frac{1324}{105}\approx12.6095$$
Which is all fine and well, except that I really don't know if I did this correctly. Does it make sense that volume should be negative? And what about 12.6095 being greater than 8?
It would have been somewhat more helpful if you had written out a bit more of your process. I am getting for the first integral:
$$ \int_{-1}^{1} \ \int_{-1}^{2} \ x^2 + 2xy + y^2 \ \ dy \ dx \ = \ \int_{-1}^{1} \ (2x^2 + 4x + \frac{8}{3} + x^2 - x + \frac{1}{3} ) \ dx $$
$$= \int_{-1}^{1} \ (3x^2 + 3x + 3 ) \ dx \ = \ (x^3 + 3x)\ |_{-1}^{1} \ = \ 8 \ . $$
So I agree with you there. I removed all odd functions in $x$ in the last integration, since the interval is symmetrical about the y-axis.
The second integral is
$$\int_{-1}^{1} \int^{2}_{1+x^2} \ x^2 + 2xy + y^2 \ \ dy \ dx \ = \ \int_{-1}^{1} \ (x^2y + xy^2 + \frac{1}{3}y^3) \ |^2_{1+x^2} \ dx $$
$$= \ \int_{-1}^{1} \ (2x^2 + 4x + \frac{8}{3} - x^2 - x^4 - x - 2x^3 - x^5 - \frac{1}{3} - x^2 - x^4 - \frac{1}{3}x^6) \ \ dx $$
$$= \ \int_{-1}^{1} \ ( 3x + \frac{7}{3} - 2x^4 - 2x^3 - x^5 - \frac{1}{3}x^6) \ \ dx \ = \ \frac{14}{3} - \frac{4}{5} - \frac{2}{21} \ = \ \frac{396}{105} = \frac{132}{35} \ ,$$
again removing the odd functions in $x$.
And the third integral is
$$ \int_{-1}^{1} \int_{y^2}^{1} \ x^2 + 2xy + y^2 \ \ dx \ dy \ = \ \int_{-1}^{1} \ (\frac{1}{3}x^3 + x^2y + xy^2) \ |_{y^2}^{1} \ \ dy $$
$$= \ \int_{-1}^{1} \ (\frac{1}{3} + y + y^2 - \frac{1}{3}y^6 - y^5 - y^4) \ \ dy \ = \ \frac{2}{3} + \frac{2}{3} - \frac{2}{5} - \frac{2}{21} \ = \ \frac{88}{105} \ , $$
taking out the odd functions in $y$. All the integrals are now positive (and, significantly, have even numerators, as we would expect from the symmetrical integrations), and the surface integral (which can also be read as a volume integration) yields $ 8 - \frac{132}{35} - \frac{88}{105} \ = \ \frac{356}{105} \ \approx 3.3905.$
[So I am getting all of your values, though I don't know where you're getting your minus signs. (You didn't count your second and third integrals as negative and then subtract them off again, did you?) This is one weird region and a bear of a computation...]