How find this integral $\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$ let $$D=\{(x,y)|y\ge x^3,y\le 1,x\ge -1\}$$
Find the integral
$$I=\dfrac{1}{2}\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$$
My idea:
$$I=\int_{0}^{1}dx\int_{x^3}^{1}(x^2y+2y^2)dy+\int_{-1}^{0}dx\int_{0}^{-x^3}(xy^2+2x+2y^2)dy$$
so
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2y^2+\dfrac{2}{3}y^3]|_{x^3}^{1}dx+\int_{-1}^{0}[\dfrac{1}{3}xy^3+2xy+\dfrac{2}{3}y^3]|_{0}^{-x^3}dx$$
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2+\dfrac{2}{3}-\dfrac{1}{2}x^8-\dfrac{2}{3}x^9]dx+\int_{-1}^{0}[-\dfrac{1}{3}x^{10}-2x^4-\dfrac{2}{3}x^9]dx$$
so
$$I=\dfrac{5}{6}-\dfrac{1}{18}-\dfrac{2}{30}+\dfrac{1}{33}+\dfrac{1}{10}-\dfrac{2}{30}=\dfrac{67}{90}$$
My question: my reslut is true? can you someone can use computer find it value? 
because I use Tom methods to find this reslut is 
$$\dfrac{79}{270}$$
which is true? so someone can use maple help me?Thank you
 A: It seems that you already realize that the for $D$, $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x\}$ is the same as $D = \{x^3 \leq y \leq 1\} \cap \{-1  \leq x \leq 1\}$. So, for a function $f(x,y)$, you should have
$$
\iint_D f(x,y) \, dxdy = \int_{-1}^1 \int_{x^3}^1 f(x,y)\,dxdy. 
$$
However, you seem to have split up your function $f(x,y)$ over different bounds without reason. You really should be integrating
$$
\int_{-1}^1 \int_{x^3}^1 (x^2 y + xy^2 + 2x + 2y^2)\,dxdy  = \int_{-1}^1 \left[ 
\frac{1}{2}x^2y^2 + \frac{1}{3} xy^3 + 2xy +\frac{2}{3} y^3 \right]_{y=x^3}^{y=1}\,dx= ...
$$
A: The domain $D$ looks roughly like a right triangle $ABC$ with the right angle $B$ at $(-1,1)$, $A$ at $(-1,-1)$, $C$ at $(1,1)$ and a curve $y=x^3$ instead of a straight line from $A$ to $C$. Since the curve does not do anything tricky (one value of $x$ maps to one value of $y$ and vice-versa) you can do this as a single integral, integrating either $x$ or $y$ first.
Your tactical choice of breaking the integral into two regions separated by the Y axis led you to make a simple error in the limits in the second region, which should still have been $\int_{x^3}^{1}$.  You also left out some of the terms in the integrand.
The correct answer is 
$$
I = \int_{x=-1}^{1} dx \int_{y=x^3}^{1}\left( x^2y + xy^2 + 2x + 2y^2 \right) dy
$$ 
$$
I =  \int_{x=-1}^{1} dx \left[ \frac{1}{2}y^2 x^2 + \frac{1}{3} y^3 x + 2yx + \frac{2}{3} y^3 \right]_{x^3}^1
$$
$$
I =  \int_{x=-1}^{1} dx \left[ \frac{1}{2}x^2 + \frac{1}{3}  x + 2x + \frac{2}{3} 
- \frac{1}{2}x^8 - \frac{1}{3} x^{10} - 2x^4 - \frac{2}{3}x^9 
\right]
$$
$$
I =  2 \left[ \frac{1}{6} + 0 + 0 + \frac{2}{3} 
- \frac{1}{18} - \frac{1}{33}  - \frac{2}{5}  - 0 
\right]
$$
(here, we use the fact that the integral from $-1$ to $1$ of an odd power of $x$ is zero to drop terms with $x$ and $x^9$)
$$
I = \frac{344}{495}
$$
A: $\frac{67}{90}$ doesn't look correct. Here is what wolfram computes
