MHB Jamie's question from Yahoo Answers regarding centroids

  • Thread starter Thread starter Chris L T521
  • Start date Start date
  • Tags Tags
    Centroids
Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
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.
 
Mathematics news on Phys.org
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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top