Find the area of the region enclosed between | vedclass

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

Question

(A) The equations of the given circles are:$x^{2}+y^{2}=4$  ........ $(1)$$(x-2)^{2}+y^{2}=4$  ........ $(2)$Equation $(1)$ is a circle with center $O

Vedclass · Accepted Answer

$\frac{8 \pi}{3}-2 \sqrt{3}$

Vedclass · Answer

(A) The equations of the given circles are:$x^{2}+y^{2}=4$  ........ $(1)$$(x-2)^{2}+y^{2}=4$  ........ $(2)$Equation $(1)$ is a circle with center $O(0,0)$ and radius $2$. Equation $(2)$ is a circle with center $C(2,0)$ and radius $2$.Solving equations $(1)$ and $(2)$:$(x-2)^{2}+y^{2}=x^{2}+y^{2}$$x^{2}-4x+4+y^{2}=x^{2}+y^{2}$$4x=4 \implies x=1$Substituting $x=1$ in $(1)$,$1+y^{2}=4 \implies y^{2}=3 \implies y=\pm \sqrt{3}$.The points of intersection are $A(1, \sqrt{3})$ and $A'(1, -\sqrt{3})$.The required area is symmetric about the $x$-axis,so area $= 2 	imes 	ext{Area}(OACA'O 	ext{ in the first quadrant})$.Area $= 2 \left[ \int_{0}^{1} \sqrt{4-(x-2)^{2}} dx + \int_{1}^{2} \sqrt{4-x^{2}} dx ight]$Using the formula $\int \sqrt{a^{2}-x^{2}} dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$:Area $= 2 \left[ \left( \frac{x-2}{2}\sqrt{4-(x-2)^{2}} + 2\sin^{-1}(\frac{x-2}{2}) ight)_{0}^{1} + \left( \frac{x}{2}\sqrt{4-x^{2}} + 2\sin^{-1}(\frac{x}{2}) ight)_{1}^{2} ight]$$= 2 \left[ \left( (-\frac{1}{2}\sqrt{3} + 2\sin^{-1}(-\frac{1}{2})) - (-\sqrt{3} + 2\sin^{-1}(-1)) ight) + \left( (0 + 2\sin^{-1}(1)) - (\frac{1}{2}\sqrt{3} + 2\sin^{-1}(\frac{1}{2})) ight) ight]$$= 2 \left[ (-\frac{\sqrt{3}}{2} - \frac{\pi}{3} + \sqrt{3} + \pi) + (\pi - \frac{\sqrt{3}}{2} - \frac{\pi}{3}) ight]$$= 2 \left[ \frac{4\pi}{3} - \sqrt{3} ight] = \frac{8\pi}{3} - 2\sqrt{3}$.

Find the area of the region enclosed between the two circles $:$ $x^{2}+y^{2}=4$ and $(x-2)^{2}+y^{2}=4$.

Similar Questions

The polynomial $f(x)$ satisfies the condition $f(x + 1) = x^2 + 4x$. The area enclosed by $y = f(x - 1)$ and the curve $x^2 + y = 0$ is

Find the area between the curves $y=x$ and $y=x^{2}$.

The area enclosed by the curves $y = x^2$,$y = x^3$,$x = 0$,and $x = p$,where $p > 1$,is $1/6$. The value of $p$ is:

The area of the region given by $\{(x, y): xy \leq 8, 1 \leq y \leq x^2\}$ is :

The area bounded by the curves $y = |x| - 1$ and $y = -|x| + 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module