Question

The area bounded by the curves $x^2+y^2=25,4y=\left|4-x^2\right|$ and $x=0$, above $x$ - axis is

• A. $2+\frac{25}{2}{\sin }^{-1}\frac{4}{5}$
• B. $2+\frac{25}{4}{\sin }^{-1}\frac{4}{5}$
• C. $2+\frac{25}{2}{\sin }^{-1}\frac{1}{5}$
• D. None of these

Answer 1

Answer: (A)

The required area

$=\underset{0}{\overset{4}{\int }}\sqrt{25-x^2}dx-\underset{0}{\overset{2}{\int }}\frac{4-x^2}{4}dx-\underset{2}{\overset{4}{\int }}\frac{x^2-4}{4}dx$
$={\left[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{\sin }^{-1}\frac{x}{5}\right]}_0^4-\frac{1}{4}{\left[4x-\frac{x^3}{3}\right]}_0^2-\frac{1}{4}{\left[\frac{x^3}{3}-4x\right]}_2^4$
$=2+\frac{25}{2}{\sin }^{-1}\frac{4}{5}$