The area (in square units) of the region encl | vedclass

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

Question

(D) The intersection points of the two circles $x^2+y^2=1$ $(i)$ and $(x-1)^2+y^2=1$ (ii) are found by substituting $y^2=1-x^2$ from $(i)$ into (ii):$

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The intersection points of the two circles $x^2+y^2=1$ $(i)$ and $(x-1)^2+y^2=1$ (ii) are found by substituting $y^2=1-x^2$ from $(i)$ into (ii):$(x-1)^2+(1-x^2)=1$$x^2-2x+1+1-x^2=1$$-2x+2=1 \Rightarrow x=\frac{1}{2}$Substituting $x=\frac{1}{2}$ into $(i)$,we get $y^2=1-\frac{1}{4}=\frac{3}{4} \Rightarrow y=\pm\frac{\sqrt{3}}{2}$.The intersection points are $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}ight)$ and $C\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}ight)$.The area of the enclosed region is symmetric about the $x$-axis,so the total area is $2 	imes$ (Area of the region in the first quadrant bounded by $x=0, x=\frac{1}{2}, x=1$ and the arcs of the circles).Area $= 2 \left[ \int_0^{1/2} \sqrt{1-(x-1)^2} dx + \int_{1/2}^1 \sqrt{1-x^2} dx ight]$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-1}{2}\sqrt{1-(x-1)^2} + \frac{1}{2}\sin^{-1}(x-1) ight)_0^{1/2} + \left( \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) ight)_{1/2}^1 ight]$$= 2 \left[ \left( \frac{-1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(-\frac{1}{2}) - (0 + \frac{1}{2}\sin^{-1}(-1)) ight) + \left( (0 + \frac{1}{2}\sin^{-1}(1)) - (\frac{1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(\frac{1}{2})) ight) ight]$$= 2 \left[ -\frac{\sqrt{3}}{8} - \frac{\pi}{12} + \frac{\pi}{4} + \frac{\pi}{4} - \frac{\sqrt{3}}{8} - \frac{\pi}{12} ight]$$= 2 \left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{4} ight] = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$.

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

Similar Questions

Let $S$ be the region bounded by the curves $y=x^{3}$ and $y^{2}=x$. The curve $y=2|x|$ divides $S$ into two regions of areas $R_{1}$ and $R_{2}$. If $\max \{R_{1}, R_{2}\}=R_{2}$,then $\frac{R_{2}}{R_{1}}$ is equal to

The area (in sq. units) of the region $A = \{(x, y) : x^2 \le y \le x + 2\}$ is

If the area between $y=m x^{2}$ and $x=m y^{2}$ $(m>0)$ is $1/4$ sq units,then the value of $m$ is

Find the area bounded by the curves $y = x^2$ and $y = 2 - x^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module