Science:Math Exam Resources/Courses/MATH101/April 2011/Question 03 (b)

MATH101 April 2011

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Question 03 (b)
Find the centroid of the finite region in the plane bounded by the x-axis and the curve $\displaystyle {y=1+\sin x}$, for $-{\frac {\pi }{2}}\leq x\leq {\frac {\pi }{2}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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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