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

As suggested in the hint given in the question, we will first solve for ${\textstyle y}$ in the equation defining the boundary of the region. After subtracting ${\textstyle x^{2}}$ from both sides, we take square roots and get

$${\frac {5y}{4}}-{\sqrt {|x|}}=\pm {\sqrt {1-x^{2}}}.$$

Then we add ${\textstyle {\sqrt {|x|}}}$ to both sides and multiply through by ${\textstyle 4/5}$ to obtain

$$y={\frac {4}{5}}\left({\sqrt {|x|}}\pm {\sqrt {1-x^{2}}}\right).$$

The region ${\textstyle {\mathcal {H}}}$ is the region above the curve ${\textstyle y={\frac {4}{5}}\left({\sqrt {|x|}}-{\sqrt {1-x^{2}}}\right)}$ and below the curve ${\textstyle y={\frac {4}{5}}\left({\sqrt {|x|}}+{\sqrt {1-x^{2}}}\right)}$. Note that these two curves meet when ${\textstyle x=\pm 1}$. Thus, the area of ${\textstyle {\mathcal {H}}}$ is given by

$$\int _{-1}^{1}{\frac {4}{5}}\left({\sqrt {|x|}}+{\sqrt {1-x^{2}}}\right)-{\frac {4}{5}}\left({\sqrt {|x|}}-{\sqrt {1-x^{2}}}\right)\;dx={\frac {8}{5}}\int _{-1}^{1}{\sqrt {1-x^{2}}}\;dx={\frac {4}{5}}{\Big [}{\sqrt {1-x^{2}}}x+\arcsin(x){\Big ]}_{-1}^{1}={\frac {8}{5}}\arcsin(1)={\frac {4\pi }{5}}.$$

Answer: The correct answer is ${\textstyle {\color {blue}{\frac {4\pi }{5}}}}$.