Question

The area of the region bounded by $y^2=16-x^2,y=0,x=0$ in the first quadrant is (in square units)

• A. $8\pi$
• B. $6\pi$
• C. $2\pi$
• D. $4\pi$
• E. $\frac{\pi }{2}$

Answer 1

The correct option is D $4\pi$

We, $y^2=16-x^2$
$\Rightarrow x^2+y^2=\mathrm{16....}\left(i\right)$
Eq. (i) is equation of a circle with radius $4$.
Required area $={\int }_0^4\sqrt{16-x^2}dx$
$={(\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4})}_0^4=4\pi$