How do I evaluate this double-integral? The problem statement is as follows:
Evaluate $\iint xydydx$ over the region bounded by the curves $x = y^2$, $x = 2 - y$, $y = 0$ and $y = 1$. 
My confusion is with regards to how the question is to be interpreted. I have tried the following two methods:


*

*Plot the curves on the graph and find the coordinates of the points enclosing the region of interest - doing so, I obtained the limits: $x$ goes from 0 to 2 (derived from the plot), $y$ goes from 0 to 1 (as stated in the question). Thus, my integral would look like so:
$$\int_0^1 \int_{0}^{2}xy dydx$$ and evaluate to 1.





*Treat the integral like so:$$\int_0^1 \int_{y^2}^{2-y}xy dydx$$ in which case, the integral evaluates to: $$\frac 14[4 + y^2 - 4y - y^4]$$


Which of these is the right way to go and how do I identify what is expected of me in solving the question?
Added a picture of my working.
 A: It is a good idea to draw the region https://www.desmos.com/calculator/7v8tu9svjk
Your second attempt is good 
\begin{eqnarray*}
\int_0^1 y \left( \int_{y^2}^{2-y} x dx \right) dy .
\end{eqnarray*}
Now the inner integral gives
\begin{eqnarray*}
 \int_{y^2}^{2-y} x dx =  \left[ \frac{x^2}{2} \right]_{y^2}^{2-y}.
\end{eqnarray*}
I will let you complete the problem from here.
Edit : This will give 
\begin{eqnarray*}
 \int_0^1  \frac{y}{2} \left[ 4-4y+y^2-y^4 \right]dy
\end{eqnarray*}
& Don Antonio's value $\color{red}{\frac{3}{8}}$ agrees with Wolfram alpha http://www.wolframalpha.com/input/?i=int_0%5E1(+int_%7By%5E2%7D%5E%7B2-y%7D+x+y+dx)+dy and the value I have calculated in my notebook.
A: Do a drawing of the horizontal parabola and the straight lines. We have:
$$\int_0^1\int_{y^2}^1xy\,\mathrm dx\,\mathrm dy+\int_1^2\int_0^{2-x}xy\,\mathrm dy\,\mathrm dx=\frac12\int_0^1y(1-y^4)\,dy+\frac12\int_1^2x\left((2-x)^2-0^2\right)\,dx$$
$$=\frac14-\frac1{12}+\frac12\int_1^2\left(4x-4x^2+x^3\right)\,dx=\frac16+\frac12\left(2(4-1)-\frac43(8-1)+\frac14(16-1)\right)=$$
$$=\frac16+3-\frac{28}6+\frac{15}8=\frac{4+72-112+45}{24}=\frac9{24}=\frac38$$
Observe that your integration domain is irregular and thus we had to split it into two regular regions...Check the arithmetic above.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[#ffd,10px]{\ds{\iint_{\large\mathbb{R}^{2}}xy\bracks{x > 0}\bracks{y > 0}
\bracks{y < 2 - x}\bracks{y < \root{x}}\dd x\,\dd y}}
\\[5mm] = &\
\int_{0}^{\infty}y\int_{0}^{\infty}x\bracks{y^{2} < x < 2 - y}\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{\infty}y\bracks{y^{2} < 2 - y}\int_{y^{\large 2}}^{2 - y}x\,\dd x\,\dd y =
\int_{0}^{\infty}y\bracks{-2 < y < 1}
\pars{2 - 2y + {1 \over 2}\,y^{2} - {1 \over 2}\,y^{4}}\dd y
\\[5mm] = &\
\int_{0}^{1}\pars{2y - 2y^{2} + {1 \over 2}\,y^{3} - {1 \over 2}\,y^{5}}\dd y = 1 - {2 \over 3} + {1 \over 8} - {1 \over 12} =
\bbx{3 \over 8} = 0.375
\end{align}

$\ds{\bracks{\cdots}}$ is an Iverson Bracket which is very helpful to handle constraints: $\ds{\bracks{P} = \color{red}{1}}$ whenever $\ds{P}$ is true and $\ds{\color{red}{0}}$ otherwise.

