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

MATH101 April 2018

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Question 11 (i)
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). (i) Determine explicitly the area of ${\textstyle {\mathcal {H}}}$ . Hint: First solve the expression in the equation above for two choices of ${\textstyle y}$ without expanding out the terms in the brackets.

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Hint
The area of a region below a curve ${\textstyle y=f(x)}$ and above a curve ${\textstyle y=g(x)}$ between ${\textstyle x=a}$ and ${\textstyle x=b}$ is given by ${\textstyle \int _{a}^{b}f(x)-g(x)\;dx}$ .

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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. 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 ${\textstyle {\color {blue}{\frac {4\pi }{5}}}}$ .