Area of the region {(x, y): x^2+(y-2)^2 leq 4 | vedclass

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Question

(D) The region is bounded by the circle $x^2 + (y-2)^2 = 4$ (center $(0, 2)$,radius $2$) and the parabola $x^2 = 2y$ (vertex $(0, 0)$ opening upwards)

Vedclass · Accepted Answer

$2\pi + \frac{16}{3}$

Vedclass · Answer

(D) The region is bounded by the circle $x^2 + (y-2)^2 = 4$ (center $(0, 2)$,radius $2$) and the parabola $x^2 = 2y$ (vertex $(0, 0)$ opening upwards).To find the intersection points,substitute $x^2 = 2y$ into the circle equation:$2y + (y-2)^2 = 4$$2y + y^2 - 4y + 4 = 4$$y^2 - 2y = 0 \implies y(y-2) = 0$So,$y = 0$ or $y = 2$.For $y = 2$,$x^2 = 4 \implies x = \pm 2$. The intersection points are $(2, 2)$ and $(-2, 2)$.The area is given by the integral of the upper curve minus the lower curve between $x = -2$ and $x = 2$.Area $= \int_{-2}^{2} [(\sqrt{4 - x^2} + 2) - \frac{x^2}{2}] dx$$= 2 \int_{0}^{2} (\sqrt{2^2 - x^2} + 2 - \frac{x^2}{2}) dx$$= 2 [(\frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}(\frac{x}{2})) + 2x - \frac{x^3}{6}]_0^2$$= 2 [(0 + 2\sin^{-1}(1)) + 4 - \frac{8}{6}]$$= 2 [2(\frac{\pi}{2}) + 4 - \frac{4}{3}]$$= 2 [\pi + \frac{8}{3}] = 2\pi + \frac{16}{3}$.

Area of the region $\{(x, y): x^2+(y-2)^2 \leq 4, x^2 \geq 2y\}$ is

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