The area of the region,enclosed by the circle | vedclass

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

Question

(B) The circle is given by $x^{2}+y^{2}=2$,so its radius $r = \sqrt{2}$. The area of the circle is $\pi r^{2} = \pi(\sqrt{2})^{2} = 2\pi$.The region c

Vedclass · Accepted Answer

$\frac{1}{6}(12 \pi-1)$

Vedclass · Answer

(B) The circle is given by $x^{2}+y^{2}=2$,so its radius $r = \sqrt{2}$. The area of the circle is $\pi r^{2} = \pi(\sqrt{2})^{2} = 2\pi$.The region common to the parabola $y^{2}=x$ and the line $y=x$ is bounded by the points of intersection. Setting $y^{2}=x$ and $y=x$ gives $x^{2}=x$,so $x(x-1)=0$,meaning $x=0$ and $x=1$. The points of intersection are $(0,0)$ and $(1,1)$.The area $A$ of the region bounded by the parabola and the line is given by the integral:$A = \int_{0}^{1} (\sqrt{x} - x) dx$$A = \left[ \frac{2}{3}x^{3/2} - \frac{x^{2}}{2} ight]_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$.The required area is the area of the circle minus the common area $A$:$	ext{Required Area} = 2\pi - \frac{1}{6} = \frac{12\pi - 1}{6} = \frac{1}{6}(12\pi - 1)$.

The area of the region,enclosed by the circle $x^{2}+y^{2}=2$ which is not common to the region bounded by the parabola $y^{2}=x$ and the straight line $y=x$,is

Similar Questions

The area enclosed by the curves $y = \ln x$,$y = \ln |x|$,$y = |\ln x|$,and $y = |\ln |x||$ is equal to:

Let the area of the region bounded by the curve $y=\max\{\sin x, \cos x\}$, lines $x=0, x=\frac{3\pi}{2}$ and the x-axis be $A$. Then, $A+A^{2}$ is equal to:

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 . . . . . . .

If $(\alpha, \beta)$ is the stationary point of the curve $y=2x-x^2$,then the area bounded by the curves $y=2^x, y=2x-x^2, x=0$ and $x=\alpha$ is

The area bounded by the curve $y=x^2+3$,$y=x$,$x=3$ and the $y$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module