Science:Math Exam Resources/Courses/MATH101/April 2018/Question 11 (ii)

MATH101 April 2018

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Question 11 (ii)
Since we all love calculus, lastly we will consider the heart-shaped region ${\textstyle {\mathcal {H}}}$ bounded by $\displaystyle x^{2}+\left({\frac {5y}{4}}-{\sqrt {\|x\|}}\right)^{2}=1\,.$ The plot of ${\textstyle {\mathcal {H}}}$ is shown below (where the coordinate axes have been deliberately removed). (ii) Assume that the density of the heart-shaped region is constant. Show that the ${\textstyle y}$ -coordinate of the centroid of this region, labelled by ${\textstyle {\bar {y}}}$ , can be written in the form ${\textstyle {\bar {y}}=b\int _{0}^{1}{\sqrt {x-x^{3}}}\,dx}$ . Determine the constant ${\textstyle b}$ explicitly.

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Hint
For the region below a curve $y=f(x)$ and above a curve $y=g(x)$ on the interval $[a,b]$ , the $y$ -coordinate of the centroid is given by ${\bar {y}}={\frac {1}{A}}\int _{a}^{b}\left[{\frac {f(x)+g(x)}{2}}\right][f(x)-g(x)]\;dx,$ where $A$ is the area of the region (this formula can be found in the CLP-2 Integral Calculus textbook).

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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. We will use the formula from the hint with ${\textstyle f(x)={\frac {4}{5}}({\sqrt {\|x\|}}+{\sqrt {1-x^{2}}})}$ and ${\textstyle g(x)={\frac {4}{5}}({\sqrt {\|x\|}}-{\sqrt {1-x^{2}}})}$ and ${\textstyle a=-1,b=1}$ ; these were obtained already in part (i). Then ${\textstyle f(x)+g(x)={\frac {8}{5}}{\sqrt {\|x\|}}}$ and ${\textstyle f(x)-g(x)={\frac {8}{5}}{\sqrt {1-x^{2}}}}$ . Moreover, we know from part (i) that ${\textstyle A=4\pi /5}$ . Thus, the ${\textstyle y}$ -coordinate of the centroid of ${\textstyle {\mathcal {H}}}$ is precisely ${\overline {y}}={\frac {1}{{\frac {4}{5}}\pi }}\int _{-1}^{1}{\frac {{\frac {8}{5}}{\sqrt {\|x\|}}}{2}}\cdot {\frac {8}{5}}{\sqrt {1-x^{2}}}\;dx={\frac {8}{5\pi }}\int _{-1}^{1}{\sqrt {\|x\|(1-x^{2})}}\;dx.$ Since the integrand is an even function, we have ${\frac {8}{5\pi }}\int _{-1}^{1}{\sqrt {\|x\|(1-x^{2})}}=2\cdot {\frac {8}{5\pi }}\int _{0}^{1}{\sqrt {\|x\|(1-x^{2})}}\;dx={\frac {16}{5\pi }}\int _{0}^{1}{\sqrt {x-x^{3}}}\;dx,$ using the fact that ${\textstyle \|x\|=x}$ for ${\textstyle x\geq 0}$ . Answer: The correct constant is ${\textstyle {\color {blue}b={\frac {16}{5\pi }}}}$ .