The area bounded by x^2 + y^2 - 2x = 0 and y | vedclass

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

Question

(A) The given circle is $x^2 + y^2 - 2x = 0$,which can be written as $(x-1)^2 + y^2 = 1^2$. This is a circle with center $(1, 0)$ and radius $r = 1$.T

Vedclass · Accepted Answer

$\frac{\pi}{2} - \frac{4}{\pi}$

Vedclass · Answer

(A) The given circle is $x^2 + y^2 - 2x = 0$,which can be written as $(x-1)^2 + y^2 = 1^2$. This is a circle with center $(1, 0)$ and radius $r = 1$.The area of the upper half of the circle is $\frac{1}{2} \pi r^2 = \frac{\pi}{2}$.The area bounded by the curve $y = \sin \frac{\pi x}{2}$ and the $x$-axis from $x = 0$ to $x = 2$ is given by $\int_{0}^{2} \sin \frac{\pi x}{2} dx$.Evaluating the integral: $\int_{0}^{2} \sin \frac{\pi x}{2} dx = \left[ -\frac{2}{\pi} \cos \frac{\pi x}{2} \right]_{0}^{2} = -\frac{2}{\pi} (\cos \pi - \cos 0) = -\frac{2}{\pi} (-1 - 1) = \frac{4}{\pi}$.The area bounded by the circle and the curve in the upper half is the area of the upper semicircle minus the area under the sine curve: $\frac{\pi}{2} - \frac{4}{\pi}$ square units.

The area bounded by $x^2 + y^2 - 2x = 0$ and $y = \sin \frac{\pi x}{2}$ in the upper half of the circle is:

Similar Questions

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 . . . . . . .

Find the area (in square units) of the region bounded by the circle $x^2 + y^2 = 9$ and the parabola $y^2 \leq 8x$.

The area bounded by the curves $y=2x^2$, $y=\max \{x-[x], x+|x|\}$ and the lines $x=0, x=2$ (in sq units) is

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

If the area of the region bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$ is equal to $\alpha$,then $6 \alpha$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module