How to find the bounds for a double integral over a triangle?

Question

I'm having trouble with this problem. I drew out the triangle and calculated the slopes to find the bounds of $x$, and the bounds of $y$ are pretty much given $0\rightarrow 13$. So I evaluate this double integral as follows and that's my answer, but it seems to be wrong. Are my bounds wrong?

$$\int _0^{13}\:\int _{\left(\frac{4}{13}y\right)}^{\left(\frac{6}{13}y\right)}\:\left(4-8x\right)dxdy$$

Answer 1

You split the integration for the intervals $x\in[0,4]$ and $x\in[4,6]$, see this for a sketch.

Thus $$\int\int_D(4-8x)dA$$ $$=\int_{0}^{4}\int_{\frac{13}{6}x}^{\frac{13}{4}x}(4-8x)dydx+\int_{4}^{6}\int_{\frac{13}{6}x}^{13}(4-8x)dydx$$

---

Splitting the integral is necessary since we have $\frac{13}{6}x\leq y\leq \frac{13}{4}x$ for $0\leq x\leq4$ and $\frac{13}{4}x\leq y\leq 13$ for $4\leq x\leq6$, so $$\int_{0}^{4}\int_{\frac{13}{6}x}^{\frac{13}{4}x}(4-8x)dydx+\int_{4}^{6}\int_{\frac{13}{6}x}^{13}(4-8x)dydx$$ $$=\int_{0}^{4}\frac{13}{12}x(4-8x)dx+\int_{4}^{6}(4-8x)(13-\frac{13}{6}x)dx$$ $$=-\frac{1352}{9}-\frac{1300}{9}=-\frac{884}{3}$$

Answer 2

I think your calculation is off. The result should be $-\frac{884}{3}$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\iint_{\mathbb{R}^{2}} \pars{\color{red}{4 - 8x}} \braces{\bracks{0 < x < 4}\bracks{y < {13 \over 4}x} + \bracks{4 < x < 6}\bracks{y < 13}}} \\ &\ \bbox[5px,#ffd]{\bracks{y > {13 \over 6}x}\dd x\,\dd y} \\[5mm] = &\ \iint_{\mathbb{R}^{2}} \braces{\bracks{0 < x < 4}\bracks{{13 \over 6}x < y < {13 \over 4}x} + \bracks{4 < x < 6}\bracks{{13 \over 6}x < y < 13}} \\ &\ \pars{\color{red}{4 - 8x}}\dd y\,\dd x \\[5mm] = &\ \int_{0}^{4}\pars{4 - 8x}\pars{{13 \over 4}x - {13 \over 6}x}\dd x + \int_{4}^{6}\pars{4 - 8x}\pars{13 - {13 \over 6}x}\dd x \\[5mm] = &\ \bbx{\large -\,{884 \over 3}} \approx -294.6667 \\ & \end{align}