# The area of the region described by A={$(x,y):(x^2+y^2\leq 1$ and $y^2\leq 1-x$}

$\begin{array}{1 1}(A)\;\large\frac{\pi}{2}+\frac{4}{3}\\(B)\;\large\frac{\pi}{2}-\frac{4}{3}\\(C)\;\large\frac{\pi}{2}-\frac{2}{3}\\(D)\;\large\frac{\pi}{2}+\frac{2}{3}\end{array}$

On solving the equation
$x^2 + y^2 = 1$ and $y^2 = 1 -x$
we get the point of induction as (1,0) and (0,1) (0, -1)
Hence the required area is
=\begin{align*} \int_0^1 \sqrt{1 -x^2} dx + \int_0^1 \sqrt{1-x} dx \end{align*}
$=\begin{bmatrix}\frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1} (x)\end{bmatrix}_0^1+ \begin{bmatrix} \frac{-(1-x)^{\frac{3}{2}}}{\frac{3}{2} } \end{bmatrix}_0^1$
$= 2 \times \begin{bmatrix} \frac{\pi}{2} \times \frac{1}{2} \end{bmatrix} + 2 \times \frac{2}{3}$
$= \frac{\pi}{2} + \frac{4}{3}$ sq.units
edited Nov 7