Bounds for a double integral 
*

*I have to integrate $\mathrm{f}\left(x,y\right) \equiv x^{2} + y^{2}$ over a triangular region with vertices $\left(0,0\right), \left(1,0\right)\ \mbox{and}\ \left(0,1\right)$.

*For my upper bound when integrating in respect to $y$, I got $1 - 1x$. Is this correct, and wouldn't this be the upper bound when integrating in respect to x as well ?. 

 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[10px,#ffd]{\int_{0}^{1}\int_{0}^{1 - y}\pars{x^{2} + y^{2}}
\dd x\,\dd y} =
\int_{0}^{1}\pars{{1 \over 3} -y + 2y^{2} - {4 \over 3}\,y^{3}}\dd y =
\bbx{1 \over 6}
\end{align}

Another way:
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}\int_{0}^{1}\bracks{x + y < 1}
\pars{x^{2} + y^{2}}\dd x\,\dd y} =
2\int_{0}^{1}\int_{0}^{1}\bracks{x + y < 1}x^{2}\,\dd x\,\dd y
\\[5mm] = &\
2\int_{0}^{1}\int_{0}^{1 - y}x^{2}\,\dd x\,\dd y =
2\int_{0}^{1}\int_{0}^{y}x^{2}\,\dd x\,\dd y =
2\int_{0}^{1}{y^{3} \over 3}\,\dd y
\\[5mm] = &\
{2 \over 3}\times {1 \over 4} = \bbx{1 \over 6}
\end{align}
