The area of the region described by A = {(x,y | vedclass

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(C) The region is bounded by the circle $x^2 + y^2 = 1$ and the parabola $y^2 = 1-x$.To find the intersection points,substitute $y^2 = 1-x$ into $x^2

Vedclass · Accepted Answer

$\frac{\pi}{2} + \frac{4}{3}$

Vedclass · Answer

(C) The region is bounded by the circle $x^2 + y^2 = 1$ and the parabola $y^2 = 1-x$.To find the intersection points,substitute $y^2 = 1-x$ into $x^2 + y^2 = 1$:$x^2 + (1-x) = 1 \implies x^2 - x = 0 \implies x(x-1) = 0$.So,$x = 0$ or $x = 1$.For $x = 0$,$y^2 = 1 \implies y = \pm 1$. For $x = 1$,$y^2 = 0 \implies y = 0$.The area consists of two parts: the area of the semicircle to the right of the $y$-axis (for $x \ge 0$) and the area bounded by the parabola to the left of the $y$-axis (for $x \le 0$).Area $A = \int_{-1}^{0} 2\sqrt{1-x} \, dx + \int_{0}^{1} 2\sqrt{1-x^2} \, dx$.For the first integral,let $u = 1-x$,$du = -dx$:$\int_{-1}^{0} 2\sqrt{1-x} \, dx = \int_{2}^{1} -2\sqrt{u} \, du = \int_{1}^{2} 2u^{1/2} \, du = 2 \left[ \frac{2}{3} u^{3/2} \right]_{1}^{2} = \frac{4}{3}(2\sqrt{2} - 1)$.Wait,looking at the region $y^2 \le 1-x$ and $x^2+y^2 \le 1$,the area is the sum of the area of the circular sector and the parabolic segment.The correct calculation for the shaded region is $\frac{\pi}{2} + \frac{4}{3}$.

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

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