Science:Math Exam Resources/Courses/MATH101/April 2018/Question 11 (ii)/Solution 1

We will use the formula from the hint with $f(x)=\frac{4}{5}(\sqrt{|x|}+\sqrt{1-x^2})$ and $g(x)=\frac{4}{5}(\sqrt{|x|}-\sqrt{1-x^2})$ and $a=-1,b=1$; these were obtained already in part (i).

Then $f(x)+g(x)=\frac{8}{5}\sqrt{|x|}$ and $f(x)-g(x)=\frac{8}{5}\sqrt{1-x^2}$. Moreover, we know from part (i) that $A=4\pi /5$. Thus, the $y$-coordinate of the centroid of ${\mathscr{H}}$ is precisely $$\bar{y}=\frac{1}{\frac{4}{5}\pi }\underset{-1}{\overset{1}{\int }}\frac{\frac{8}{5}\sqrt{|x|}}{2}\cdot \frac{8}{5}\sqrt{1-x^2}dx=\frac{8}{5\pi }\underset{-1}{\overset{1}{\int }}\sqrt{|x|(1-x^2)}dx.$$ Since the integrand is an even function, we have $$\frac{8}{5\pi }\underset{-1}{\overset{1}{\int }}\sqrt{|x|(1-x^2)}=2\cdot \frac{8}{5\pi }\underset{0}{\overset{1}{\int }}\sqrt{|x|(1-x^2)}dx=\frac{16}{5\pi }\underset{0}{\overset{1}{\int }}\sqrt{x-x^3}dx,$$ using the fact that $|x|=x$ for $x\ge 0$.

Answer: The correct constant is $b=\frac{16}{5\pi }$.