Science:Math Exam Resources/Courses/MATH101/April 2018/Question 06 (ii)

MATH101 April 2018

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Question 06 (ii)
Calculate the centroid of the semi-circular region ${\textstyle x^{2}+y^{2}=R^{2}}$ with ${\textstyle y\geq 0}$, where ${\textstyle R}$ is the radius of the semi-circular region. Assume that the density of the region is constant. Hint: first draw a picture.

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
Assume that the region is given by $$\{(x,y):a\leq x\leq b,\quad B(x)\leq y\leq T(x)\}$$ for some function ${\textstyle B}$ and ${\textstyle T}$. Then, the centroid is $${\bar {x}}={\frac {\int _{a}^{b}x(T(x)-B(x))dx}{A}},\quad {\bar {y}}={\frac {\int _{a}^{b}T(x)^{2}-B(x)^{2}dx}{2A}},$$ where the area ${\textstyle A}$ is ${\textstyle A=\int _{a}^{b}T(x)-B(x)dx}$.

Hint 2
Observe that the graph of the region has a symmetry with respect to $y$-axis.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Drawing the given region, we can recognize that the functions ${\textstyle b}$ and ${\textstyle t}$ in Hint are $$B(x)=0,\quad T(x)={\sqrt {R^{2}-x^{2}}}.$$ Also, ${\textstyle a}$ and ${\textstyle b}$ are ${\textstyle a=-R}$ and ${\textstyle b=R}$. First, we find the area ${\textstyle A}$ in the Hint. Since ${\textstyle A}$ represents the area of the region and the given region is the half of the circle with a radius ${\textstyle R}$, we have $$A={\frac {1}{2}}\pi R^{2}.$$ Now, we find the centroid based on the formula in Hint. The ${\textstyle y}$-coordinate of the centroid is $${\begin{aligned}{\bar {y}}&={\frac {\int _{a}^{b}T(x)^{2}-B(x)^{2}dx}{2A}}={\frac {1}{\pi R^{2}}}\int _{-R}^{R}({\sqrt {R^{2}-x^{2}}})^{2}dx={\frac {1}{\pi R^{2}}}\int _{-R}^{R}(R^{2}-x^{2})dx\\&={\frac {2}{\pi R^{2}}}\int _{0}^{R}(R^{2}-x^{2})dx={\frac {2}{\pi R^{2}}}\left[R^{2}x-{\frac {1}{3}}x^{3}\right]_{x=0}^{x=R}\\&={\frac {2}{\pi R^{2}}}\left(R^{3}-{\frac {1}{3}}R^{3}\right)={\frac {2}{\pi R^{2}}}\cdot {\frac {2}{3}}R^{3}={\frac {4R}{3\pi }}.\end{aligned}}$$ In the forth equality, we use that the integrand is an even function. On the other hand, the ${\textstyle x}$-coordinate of the centroid is $0$. We can easily see this from its formula observing that the integrand is an odd function and the interval of the integral is $(-R,R)$: $${\begin{aligned}{\bar {x}}={\frac {\int _{a}^{b}x(T(x)-B(x))dx}{A}}={\frac {2}{\pi R^{2}}}\int _{-R}^{R}x{\sqrt {R^{2}-x^{2}}}dx=0.\end{aligned}}$$ We can also see ${\bar {x}}=0$ from the graph of the region, considering the symmetry with respect to $y$-axis. To summarize the centroid is $\left(0,{\frac {4R}{3\pi }}\right)$. Answer: $\color {blue}\left(0,{\frac {4R}{3\pi }}\right)$