Science:Math Exam Resources/Courses/MATH101/April 2008/Question 07 (b)

MATH101 April 2008

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Question 07 (b)
Let E be the ellipse x2 + k2y2 = 1, where k is a constant with k ≥ 1. Let S be the region inside the circle x2 + y2 = 1, outside E, and above the x-axis. Find all values of k such that the centroid (centre of mass) of S lies inside S (i.e. outside E). You may use the result of part (a).

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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. A picture when k is 2 is drawn below. To compute the area of S, we actually don't need to do any more work. Notice that the area of the blue circle when y is greater than 0 is half the area of the unit circle which is ${\tfrac {\pi }{2}}$. Now, the area of the ellipse when y is greater than 0 is again half the area of the ellipse, which by the previous problem is ${\tfrac {\pi (1)}{2}}({\tfrac {1}{k}})={\tfrac {\pi }{2k}}$. Hence, the area of S is $A={\frac {\pi }{2}}-{\frac {\pi }{2k}}={\frac {\pi }{2}}\left(1-{\frac {1}{k}}\right).$ Now, notice that when x is 0, then the two points on the ellipse are $y=\pm {\tfrac {1}{k}}$. So the one lying in the upper half of the plane is $y={\tfrac {1}{k}}$. The question asks when is the centroid outside of E (so inside S). By symmetry, the x coordinate of the centroid is 0. So all we need to do is figure out when ${\bar {y}}\geq {\tfrac {1}{k}}$ (NOTE: There is a bit of ambiguity in the sense that is something on the boundary of S and E inside S or outside of E - I suspect students using greater than or equal to vs just greater than would earn full marks but it is something we should be a bit careful with). In any case, we compute when ${\begin{aligned}{\frac {1}{k}}\leq {\bar {y}}&={\frac {1}{2A}}\int _{a}^{b}(f(x)^{2}-g(x)^{2})\,dx\end{aligned}}$ Here, the upper curve is the positive half of the circle so $\displaystyle f(x)={\sqrt {1-x^{2}}}$ and the lower curve is the upper half of the ellipse so $\displaystyle g(x)={\sqrt {\frac {1-x^{2}}{k^{2}}}}.$ The picture also clearly shows that we are integrating from $-1\leq x\leq 1$ (we could have explicitly found these points of intersection as well). Thus, we have ${\begin{aligned}{\frac {1}{k}}\leq {\bar {y}}&={\frac {1}{2A}}\int _{a}^{b}(f(x)^{2}-g(x)^{2})\,dx\\&={\frac {1}{2\cdot {\frac {\pi }{2}}\left(1-{\frac {1}{k}}\right)}}\int _{-1}^{1}(1-x^{2}-{\frac {1-x^{2}}{k^{2}}})\,dx\\&={\frac {1}{\pi \cdot {\frac {k-1}{k}}}}\int _{-1}^{1}(1-x^{2})\left(1-{\frac {1}{k^{2}}}\right))\,dx\end{aligned}}$ where in the last step we factored out a $1-x^{2}$. Pulling out the terms depending only on k in the integral and using the fact that the integral is the integral of an even function over a symmetric interval yields ${\begin{aligned}{\frac {1}{k}}\leq {\bar {y}}&={\frac {1}{\pi \cdot {\frac {k-1}{k}}}}\left(1-{\frac {1}{k^{2}}}\right)2\int _{0}^{1}(1-x^{2}))\,dx\\&={\frac {2k}{\pi (k-1)}}\cdot {\frac {k^{2}-1}{k^{2}}}\left.\left(x-{\frac {x^{3}}{3}}\right)\right|_{0}^{1}\\&={\frac {2k}{\pi (k-1)}}\cdot {\frac {(k+1)(k-1)}{k^{2}}}\left(1-{\frac {1}{3}}\right)\\&={\frac {4(k+1)}{3\pi k}}\end{aligned}}$ Using the inequality derived above, we have ${\begin{aligned}{\frac {1}{k}}&\leq {\frac {4(k+1)}{3\pi k}}\\3\pi &\leq 4(k+1)\\{\frac {3\pi -4}{4}}&\leq k\end{aligned}}$ and this completes the proof.

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