MATH101 April 2011
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Question 03 (b)

Find the centroid of the finite region in the plane bounded by the xaxis and the curve $\displaystyle {y=1+\sin x}$, for ${\frac {\pi }{2}}\leq x\leq {\frac {\pi }{2}}$.

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Hint 1

The coordinates of the centroid are computed using the formulas
 $\displaystyle {{\bar {x}}={\frac {1}{A}}\int _{a}^{b}xf(x)dx}$
and
 $\displaystyle {{\bar {y}}={\frac {1}{A}}\int _{a}^{b}{\frac {1}{2}}(f(x))^{2}dx}$
where
 $A=\int _{a}^{b}f(x)\,dx$

Hint 2

 $\sin(x)^{2}={\frac {1\cos(2x)}{2}}$

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Solution

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First we compute the area A of the shape. This is given by
 $\displaystyle {\begin{aligned}A&=\int _{\pi /2}^{\pi /2}1+\sin(x)dx\\&=x{\big }_{\pi /2}^{\pi /2}\cos {\big }_{\pi /2}^{\pi /2}\\&=\pi \cos(\pi /2)+\cos(\pi /2)\\&=\pi \end{aligned}}$
The coordinates of the centroid are computed using the formulas
 $\displaystyle {{\bar {x}}={\frac {1}{\pi }}\int _{\pi /2}^{\pi /2}x(1+\sin(x))dx}$
and
 $\displaystyle {{\bar {y}}={\frac {1}{\pi }}\int _{\pi /2}^{\pi /2}{\frac {1}{2}}(1+\sin(x))^{2}dx}$
We will compute ${\bar {x}}$ first.
 $\displaystyle {\begin{aligned}{\bar {x}}&={\frac {1}{\pi }}\int _{\pi /2}^{\pi /2}x+x\sin(x)dx\\&={\frac {1}{\pi }}\int _{\pi /2}^{\pi /2}xdx+{\frac {1}{\pi }}\int _{\pi /2}^{\pi /2}x\sin(x)dx\\&=0+{\frac {1}{\pi }}\int _{\pi /2}^{\pi /2}x\sin(x)dx\end{aligned}}$
The first integral is zero because the function is odd. For the second we can use integration by parts.
 $\displaystyle {\begin{aligned}\int x\sin(x)dx&=x\cos(x)+\int \cos(x)dx\\&=x\cos(x)+\sin(x)+C\end{aligned}}$
That means if we plug in the limits for the definite integral we get
 $\displaystyle {\begin{aligned}{\bar {x}}&=1/\pi (x\cos(x)+\sin(x)){\big }_{\pi /2}^{\pi /2}\\&=1/\pi (\sin(\pi /2)\sin(\pi /2))\\&=2/\pi \end{aligned}}$
Now we compute ${\bar {y}}$
 $\displaystyle {\begin{aligned}{\bar {y}}&={\frac {1}{\pi }}\int _{\pi /2}^{\pi /2}{\frac {1}{2}}(1+\sin(x))^{2}dx\\&={\frac {1}{2\pi }}\int _{\pi /2}^{\pi /2}1+2\sin(x)+\sin(x)^{2}dx\\&={\frac {1}{2\pi }}{\Big (}\int _{\pi /2}^{\pi /2}1dx+2\int _{\pi /2}^{\pi /2}\sin(x)dx+\int _{\pi /2}^{\pi /2}\sin(x)^{2}dx{\Big )}\\&={\frac {1}{2}}+{\frac {1}{2\pi }}\int _{\pi /2}^{\pi /2}\sin(x)^{2}dx\end{aligned}}$
To compute this we note $\sin(x)^{2}={\frac {1\cos(2x)}{2}}$
 $\displaystyle {\begin{aligned}\int \sin(x)^{2}dx&=\int {\frac {1\cos(2x)}{2}}dx\\&={\frac {x}{2}}{\frac {1}{4}}\sin(2x)+C\end{aligned}}$
Hence
 $\displaystyle {\begin{aligned}{\bar {y}}&={\frac {1}{2}}+{\frac {1}{2\pi }}({\frac {x}{2}}{\frac {1}{4}}\sin(2x))_{\pi /2}^{\pi /2}\\&={\frac {3}{4}}\end{aligned}}$
Thus the centroid is $({\frac {2}{\pi }},{\frac {3}{4}})$

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