Jamie's question from Yahoo Answers regarding centroids

[Chris L T521]

Here is the question.

Centroid of x+y=2 x=y^2?

Here is a link to the question:

Centroid of x+y=2 x=y^2? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Chris L T521]

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!

 

[Chris L T521]

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).

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[Chris L T521]

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[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].}
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[alyafey22]

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 .

 

[Chris L T521]

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.