Science:Math Exam Resources/Courses/MATH101/April 2009/Question 03 (a)

MATH101 April 2009

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Question 03 (a)
Let $R$ be the finite region in the $xy$-plane bounded by $x=0$, $y=0$ and $y=cosx$. (a) Calculate the centroid of $R$.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
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 $A=\int _{a}^{b}f(x)\,dx$

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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. The first point with $x>0$ where $\cos(x)$ crosses the x-axis is at $x={\frac {\pi }{2}}$. Hence we have that $a=0$ and $b={\frac {\pi }{2}}$. Thus, we have ${\begin{aligned}A&=\int _{a}^{b}f(x)\,dx\\&=\int _{0}^{\tfrac {\pi }{2}}\cos(x)\,dx\\&=\left.\sin(x)\right|_{0}^{\tfrac {\pi }{2}}\\&=\sin({\tfrac {\pi }{2}})-\sin(0)\\&=1\end{aligned}}$ The x-coordinate of the centroid are computed using the formulas ${\begin{aligned}{\bar {x}}&={\frac {1}{A}}\int _{a}^{b}xf(x)\,dx\\&={\frac {1}{1}}\int _{0}^{\tfrac {\pi }{2}}x\cos(x)\,dx\end{aligned}}$ To solve this, we use integration by parts. Let ${\begin{aligned}u=x\quad &\quad v=\sin(x)\\du=dx\quad &\quad dv=\cos(x)dx\end{aligned}}$ Then, ${\begin{aligned}{\bar {x}}&=\int _{0}^{\tfrac {\pi }{2}}x\cos(x)\,dx\\&=\left.x\sin(x)\right|_{0}^{\tfrac {\pi }{2}}-\int _{0}^{\tfrac {\pi }{2}}\sin(x)\,dx\\&=\left.x\sin(x)\right|_{0}^{\tfrac {\pi }{2}}+\cos(x)_{0}^{\tfrac {\pi }{2}}\\&={\frac {\pi }{2}}\sin({\frac {\pi }{2}})-(0)\sin(0)+\cos({\frac {\pi }{2}})-\cos(0)\\&={\frac {\pi }{2}}-1\end{aligned}}$ Lastly, for the other coordinate, we have ${\begin{aligned}{\bar {y}}&={\frac {1}{A}}\int _{a}^{b}{\frac {1}{2}}(f(x))^{2}\,dx\\&={\frac {1}{1}}\int _{0}^{\tfrac {\pi }{2}}{\frac {1}{2}}(f(x))^{2}\,dx\\&={\frac {1}{2}}\int _{0}^{\tfrac {\pi }{2}}\cos ^{2}(x)\,dx\end{aligned}}$ Now we use the double angle formula $\cos ^{2}(x)={\frac {1+\cos(2x)}{2}}$ to see that ${\begin{aligned}{\bar {y}}&={\frac {1}{2}}\int _{0}^{\tfrac {\pi }{2}}\cos ^{2}(x)\,dx\\&={\frac {1}{2}}\int _{0}^{\tfrac {\pi }{2}}{\frac {1+\cos(2x)}{2}}\,dx\\&=\left.{\frac {1}{4}}x\right|_{0}^{\tfrac {\pi }{2}}+{\frac {1}{4}}\int _{0}^{\tfrac {\pi }{2}}\cos(2x)\,dx\\&=\left.{\frac {1}{4}}x\right|_{0}^{\tfrac {\pi }{2}}+\left.{\frac {1}{8}}\sin(2x)\right|_{0}^{\tfrac {\pi }{2}}\\&={\frac {\pi }{8}}+{\frac {1}{8}}(\sin(\pi )-\sin(0))\\&={\frac {\pi }{8}}\end{aligned}}$ and so the centroid is $({\bar {x}},{\bar {y}})=\left({\frac {\pi }{2}}-1,{\frac {\pi }{8}}\right)$ completing the proof.

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