Science:Math Exam Resources/Courses/MATH101/April 2016/Question 08 (a)

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MATH101 April 2016

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Question 08 (a)
Let $A$ be the region to the right of the $y$ -axis that is bounded by the graphs of $y=x^{2}$ and $y=6-x$ . Both parts of this question concern this region $A$ . (a) Find the centroid of $A$ , assuming it has constant density $\rho =1$ . The area of $A$ is ${\frac {22}{3}}$ (you don’t have to show this). You may leave your answer in calculator-ready form.

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
Recall the formula for the centroid.

Hint 2
Find the intersection points of the graph of two given functions.

Hint 3
Find the horizontal and vertical moments.

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Solution
Setting $x^{2}=6-x$ , we see that the two curves intersect at $x=2$ (as well as $x=-3$ , which is to the left of the $y$ -axis and therefore irrelevant). The horizontal moment is ${\begin{aligned}\int _{0}^{2}x{\big (}(6-x)-x^{2}{\big )}\,dx&=\int _{0}^{2}(6x-x^{2}-x^{3})\,dx\\&={\bigg (}3x^{2}-{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}{\bigg )}{\bigg \|}_{0}^{2}\\&=12-{\frac {8}{3}}-4={\frac {16}{3}},\end{aligned}}$ while the vertical moment is ${\begin{aligned}{\frac {1}{2}}\int _{0}^{2}{\big (}(6-x)^{2}-x^{4}\ {\big )}\,dx&={\frac {1}{2}}{\bigg (}{-}{\frac {(6-x)^{3}}{3}}-{\frac {x^{5}}{5}}{\bigg )}{\bigg \|}_{0}^{2}\\&={\frac {1}{2}}{\bigg (}{\frac {216-64}{3}}-{\frac {32}{5}}{\bigg )}={\frac {332}{15}}.\end{aligned}}$ Note that the last integral could also be evaluated as ${\begin{aligned}{\frac {1}{2}}\int _{0}^{2}{\big (}36-12x+x^{2}-x^{4}\ {\big )}\,dx&={\frac {1}{2}}{\bigg (}36x-6x^{2}+{\frac {x^{3}}{3}}-{\frac {x^{5}}{5}}{\bigg )}{\bigg \|}_{0}^{2}\\&={\frac {1}{2}}{\bigg (}72-24+{\frac {8}{3}}-{\frac {32}{5}}{\bigg )}={\frac {332}{15}}.\end{aligned}}$ Using the fact that The centroid of area $A={\frac {1}{A}}({\text{The horizontal moment}},{\text{The vertical moment}}),$ the centroid is therefore $\displaystyle {\frac {1}{22/3}}{\bigg (}{\frac {16}{3}},{\frac {332}{15}}{\bigg )}={\color {blue}{\bigg (}{\frac {8}{11}},{\frac {166}{55}}{\bigg )}}$ .