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

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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:

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