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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

The area of the region lying between the curves $y=\sqrt{4-x^2}$,$y^2=3x$ and the $Y$-axis is

The area of the region bounded by the curve $y = \max \{| x |, x | x - 2 |\}$,the $x$-axis,and the lines $x = -2$ and $x = 4$ is equal to . . . . . . .

The area of the region bounded by the lines $x=1, x=2$,and the curves $x(y-e^x)=\sin x$ and $2xy=2\sin x+x^3$ is

Find the area bounded by the curves $x = \sqrt{y - 1}$ and $y = x + 1$.

Let the area of the region $\{(x, y): x-2y+4 \geq 0, x+2y^2 \geq 0, x+4y^2 \leq 8, y \geq 0\}$ be $\frac{m}{n}$,where $m$ and $n$ are coprime numbers. Then $m+n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module