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Jamie's question from Yahoo! Answers regarding centroids

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Chris L T521

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Jan 26, 2012
995
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Chris L T521

Well-known member
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Jan 26, 2012
995
Hi Jamie,

The region that we're supposed to find the centroid for can be found in the figure below.
Assuming that you know multivariable calculus, one defines the centroid as $(\bar{x},\bar{y})$ where
\[\bar{x}= \frac{\displaystyle\iint\limits_R x\,dA}{\displaystyle\iint\limits_R \,dA}=\frac{1}{\text{Area of $R$}}\iint\limits_R x\,dA\qquad\text{ and }\qquad \bar{y}= \frac{\displaystyle\iint\limits_R y\,dA}{\displaystyle\iint\limits_R \,dA}=\frac{1}{\text{Area of $R$}}\iint\limits_R y\,dA\]
Due to the region $R$ we have, it would be best to treat it as a type II region and treat the bounding functions as functions of the form $x=f(y)$ (if we treated the region as type I, then we'd have more than one integral to work with for each of $\bar{x}$ and $\bar{y}$). With that said, the two bounding functions are $x=2-y$ and $x=y^2$.

Thus, we see that
\[\begin{aligned} \iint\limits_R \,dA &= \int_{-2}^1\int_{y^2}^{2-y}\,dx\,dy \\ &= \int_{-2}^1 2-y-y^2\,dy \\ &= \left.\left[ 2y-\frac{1}{2}y^2-\frac{1}{3}y^3\right]\right|_{-2}^1\\ &= \left(2-\frac{1}{2}-\frac{1}{3}\right) - \left(-4 -2 + \frac{8}{3}\right)\\ &= \frac{9}{2}\end{aligned}\]

\[\begin{aligned} \iint\limits_R x\,dA &= \int_{-2}^1\int_{y^2}^{2-y} x\,dx\,dy\\ &= \frac{1}{2}\int_{-2}^1 \left.\left[ x^2\right]\right|_{y^2}^{2-y}\,dy\\ &= \frac{1}{2}\int_{-2}^1 4-4y+y^2-y^4\,dy \\ &= \frac{1}{2}\left.\left[ 4y - 2y^2+\frac{1}{3}y^3- \frac{1}{5}y^5\right]\right|_{-2}^1\\ &= \frac{1}{2}\left[\left( 4-2+\frac{1}{3}-\frac{1}{5}\right) - \left( -8-8-\frac{8}{3} +\frac{32}{5}\right)\right]\\ &= \frac{36}{5}\end{aligned}\]

\[\begin{aligned} \iint\limits_R y\,dA &= \int_{-2}^1\int_{y^2}^{2-y} y\,dx\,dy\\ &= \int_{-2}^1 y(2-y-y^2)\,dy\\ &= \int_{-2}^1 2y-y^2-y^3\,dy \\ &= \left.\left[y^2-\frac{1}{3}y^3-\frac{1}{4}y^4\right]\right|_{-2}^1\\ &= \left[\left(1-\frac{1}{3}-\frac{1}{4}\right) - \left(4+\frac{8}{3} - 4\right)\right]\\ &= -\frac{9}{4}\end{aligned}\]

Therefore,

\[\bar{x}= \frac{\displaystyle\iint\limits_R x\,dA}{\displaystyle\iint\limits_R \,dA}= \frac{\dfrac{36}{5}}{\dfrac{9}{2}} = \frac{8}{5}\]

and

\[\bar{y}= \frac{\displaystyle\iint\limits_R y\,dA}{\displaystyle\iint\limits_R \,dA}= \frac{-\dfrac{9}{4}}{\dfrac{9}{2}} = -\frac{1}{2}\]

Thus, the centroid of this region is $(\bar{x},\bar{y})=\left(\dfrac{8}{5},-\dfrac{1}{2}\right)$ (as seen in the figure below).
I hope this made sense!!
 
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Chris L T521

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Jan 26, 2012
995
For those of you who are interested, here's the TikZ codes for the two figures (in two separate posts because apparently I've exceeded the character limit for a single post).


Code:
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[COLOR=#800000]\centering[/COLOR]
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[COLOR=#800000]\caption[/COLOR][COLOR=#000000]{The region [/COLOR][COLOR=#008000]$R$[/COLOR][COLOR=#000000] bounded by the curves [/COLOR][COLOR=#008000]$x+y=2$[/COLOR][COLOR=#000000] and [/COLOR][COLOR=#008000]$x=y^2$[/COLOR][COLOR=#000000].}
[/COLOR][COLOR=#0000CC]\end[/COLOR][COLOR=#000000]{figure}[/COLOR]
 
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Chris L T521

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Jan 26, 2012
995

Code:
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[COLOR=#800000]\centering[/COLOR]
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[COLOR=#0000cc]\end[/COLOR][COLOR=#000000]{tikzpicture}[/COLOR]
[COLOR=#800000]\caption[/COLOR][COLOR=#000000]{The centroid of the region [/COLOR][COLOR=#008000]$R$[/COLOR][COLOR=#000000] bounded by the curves [/COLOR][COLOR=#008000]$x+y=2$[/COLOR][COLOR=#000000] and [/COLOR][COLOR=#008000]$x=y^2$[/COLOR][COLOR=#000000].}
[/COLOR][COLOR=#0000CC]\end[/COLOR][COLOR=#000000]{figure}[/COLOR]
 

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
1,667
That is really great , thanks for sharing the code . By the way , do I need a program to render the code or I can do it online ?

PS : I need to draw some contours for the lessons I am about to write .
 
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  • #6

Chris L T521

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Staff member
Jan 26, 2012
995
That is really great , thanks for sharing the code . By the way , do I need a program to render the code or I can do it online ?

PS : I need to draw some contours for the lessons I am about to write .
If you have LaTeX installed on your computer, then you should have no problem compiling it. You'll just need to include \usepackage{tikz} in the preamble of your document.