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Q.
The area enclosed between the curve $x^{2} + y^{2} = 16$ and the coordinate axes in the first quadrant is
Application of Integrals
Solution:
Given that, the circle with centre $(0,0)$ and radius $4$. Required area
$=\int\limits_{0}^{4}\sqrt{16-x^{2}} dx$
$=\left[\frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}sin^{-1}\frac{x}{4}\right]_{0}^{4}=4\pi$ sq. units
