- 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 ?.
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- $\begingroup$ $\texttt{MathJax}$ Basic Tutorial and Quick Reference. $\endgroup$Felix Marin– Felix Marin2019-03-28 21:20:22 +00:00Commented Mar 28, 2019 at 21:20
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1 Answer
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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} $\endgroup$
2 $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
- $\begingroup$ Yeah, I had gotten that answer after I read over my textbook and attempted to answer the question once more. Thanks! $\endgroup$Uchuuko– Uchuuko2019-03-28 21:28:39 +00:00Commented Mar 28, 2019 at 21:28
- $\begingroup$ @Uchuuko You're welcome. $\endgroup$Felix Marin– Felix Marin2019-03-28 21:32:26 +00:00Commented Mar 28, 2019 at 21:32