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

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 ${\textstyle {\color {blue}b={\frac {16}{5\pi }}}}$.