MHB Jamie's question from Yahoo Answers regarding centroids

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The centroid of the region defined by the equations x+y=2 and x=y^2 is calculated using multivariable calculus. The area of the region R is found to be 9/2, while the integrals for x and y yield values of 36/5 and -9/4, respectively. Consequently, the centroid coordinates are determined as (8/5, -1/2). The discussion also includes a request for information on rendering TikZ code, with confirmation that LaTeX can be used for this purpose. Overall, the thread provides a detailed mathematical analysis and practical advice for visual representation.
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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.
 
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Hi Jamie,

The region that we're supposed to find the centroid for can be found in the figure below.
mhb_centroid.png
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).
mhb_centroid2.png
I hope this made sense!
 
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).

mhb_centroid.png

Code:
\begin{figure}[!ht]
\centering
\begin{tikzpicture}[scale=.95]
   \draw[very thin,color=gray!35] (-5.5,-3.5) grid (5.5,3.5);
   \draw[<->] (-5.5,0) -- (5.5,0) node[right]{$x$};
   \draw[<->] (0,-4) -- (0,4) node[above]{$y$};
   \foreach\x in {-5,-4,-3,-2,-1,1,2,3,4,5}{
   \draw (\x,.1) -- (\x,-.1) node[below]{$\x$};
      }
   \foreach\x in {-3,-2,-1,1,2,3}{
   \draw (-.1,\x) -- (.1,\x) node[right]{$\x$};
     }
   \draw[<->,blue,thick](5.5,-3.5) -- (-1.5,3.5) node[above,left]{$x+y=2$};
   \fill.25] (1,1) -- (4,-2) -- plot[domain=0:4,smooth,samples=1500](\x,{-sqrt(\x)}) 
   -- plot[domain=0:1,smooth,samples=1500](\x,{sqrt(\x)}) --  cycle;
   \draw (1.5,-0.5) node{$R$};
   \draw[red,thick] plot[domain=0:5.5,smooth,samples=1500] (\x, {-sqrt(\x)});
   \draw[red,thick] plot[domain=0:5.5,smooth,samples=1500] (\x, {sqrt(\x)}) node[above]{$x=y^2$};
   \fill (1,1) circle (2pt) node[right=.2cm]{$(1,1)$};
   \fill (4,-2) circle (2pt) node[below=.25cm,left]{$(4,-2)$};
\end{tikzpicture}
\caption{The region $R$ bounded by the curves $x+y=2$ and $x=y^2$.}
\end{figure}
 
mhb_centroid2.png

Code:
\begin{figure}[!ht]
\centering
\begin{tikzpicture}[scale=.95]
   \draw[very thin,color=gray!35] (-5.5,-3.5) grid (5.5,3.5);
   \draw[<->] (-5.5,0) -- (5.5,0) node[right]{$x$};
   \draw[<->] (0,-4) -- (0,4) node[above]{$y$};
   \foreach\x in {-5,-4,-3,-2,-1,1,2,3,4,5}{
   \draw (\x,.1) -- (\x,-.1) node[below]{$\x$};
     }
   \foreach\x in {-3,-2,-1,1,2,3}{
   \draw (-.1,\x) -- (.1,\x) node[right]{$\x$};
     }
   \draw[<->,blue,thick](5.5,-3.5) -- (-1.5,3.5) node[above,left]{$x+y=2$};
   \fill.25] (1,1) -- (4,-2) -- plot[domain=0:4,smooth,samples=1500](\x,{-sqrt(\x)}) 
   -- plot[domain=0:1,smooth,samples=1500](\x,{sqrt(\x)}) --  cycle;
   \draw (1.5,-0.5) node{$R$};
   \draw[red,thick] plot[domain=0:5.5,smooth,samples=1500] (\x, {-sqrt(\x)});
   \draw[red,thick] plot[domain=0:5.5,smooth,samples=1500] (\x, {sqrt(\x)}) node[above]{$x=y^2$};
   \fill (1,1) circle (2pt) node[right=.2cm]{$(1,1)$};
   \fill (4,-2) circle (2pt) node[below=.25cm,left]{$(4,-2)$};
   \fill[blue] (1.6,-.5) circle (2pt);
   \draw[blue] (-2.5,-2)  node{$(\bar{x},\bar{y})=\left(\frac{8}{5},-\frac{1}{2}\right)$};
   \draw[->,color=blue] (-1,-2) to[out=0,in=270] (1.6,-.6);
\end{tikzpicture}
\caption{The centroid of the region $R$ bounded by the curves $x+y=2$ and $x=y^2$.}
\end{figure}
 
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 .
 
ZaidAlyafey said:
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.
 
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