Double Integral and probabilities 

Could someone please explain the solution to this question? I understand that I have to calculate the double integral of y1 and y2, but I don't understand why y1 is associated with the region 0 to 1/3. Shouldn't it be 0 to 1/2 for y1? I tried graphing the region and used arrows to determine the limits but I still don't get it. Any detailed explanation would be greatly appreciated. Thank you
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\mrm{P}\pars{Y_{1} \leq {1 \over 2},Y_{2} \leq {1 \over 3}} =
\int_{0}^{1}\int_{0}^{1}3y_{1}\bracks{y_{2} \leq y_{1}}
\bracks{y_{1} \leq {1 \over 2}}\bracks{y_{2} \leq {1 \over 3}}
\,\dd y_{1}\,\dd y_{2}
\\[5mm] = &\
3\int_{0}^{1/3}\int_{0}^{1/2}y_{1}\bracks{y_{1} \geq y_{2}}
\,\dd y_{1}\,\dd y_{2} =
3\int_{0}^{1/3}\bracks{y_{2} < {1 \over 2}}\int_{y_{2}}^{1/2}y_{1}
\,\dd y_{1}\,\dd y_{2}
\\[5mm] = &\
3\int_{0}^{1/3}\bracks{{1 \over 2}\,\pars{1 \over 2}^{2} -
{1 \over 2}\,y_{2}^{2}}\dd y_{2} = \bbx{\ds{23 \over 216}} \approx 0.1065 
\end{align}

The other one is quite similar to the present one.

A: Draw some axes upto the unit square & draw the line $y_1=y_2$ and shade the triangular region below this line. Firstly a quick sanity check ...
\begin{eqnarray*}
 \int_{y_2=0}^{y_1} \int_0^{1} 3y_1 dy_1  dy_2 
\end{eqnarray*}
This does indeed equal $1$ as expected. 
Now draw the horizontal line $y_2=1/3$ and the vertical line $y_1=1/2$. This gives a trapezium shape in the shaded area, split this into two parts & we have the following integrals
\begin{eqnarray*}
\int_{y_2=0}^{y_1}\int_0^{\frac{1}{3}} 3y_1 dy_1  dy_2  +\int_{y_2=0}^{\frac{1}{3}}\int_{\frac{1}{3}}^{\frac{1}{2}} 3y_1 dy_1  dy_2
\end{eqnarray*}
The first of these is $\frac{1}{27}$ & the second is $\frac{5}{72}$. Add these & we have $ \color{red}{\frac{23}{216}=0.10648 \cdots}$
