The area of the smaller part between the circ | vedclass

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(A) The equation of the circle is $x^2+y^2=2^2$,which has a radius $r=2$ and center at $(0,0)$.The line is $x=1$. The intersection points of the circl

Vedclass · Accepted Answer

$\frac{4 \pi}{3}-\sqrt{3}$

Vedclass · Answer

(A) The equation of the circle is $x^2+y^2=2^2$,which has a radius $r=2$ and center at $(0,0)$.The line is $x=1$. The intersection points of the circle and the line are found by substituting $x=1$ into the circle equation: $1^2+y^2=4 \implies y^2=3 \implies y=\pm\sqrt{3}$.The area of the smaller part is the region bounded by the circle and the line $x=1$ from $x=1$ to $x=2$.The area $A$ is given by $2 \int_{1}^{2} y \, dx = 2 \int_{1}^{2} \sqrt{4-x^2} \, dx$.Using the formula $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$,we get:$A = 2 [\frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}(\frac{x}{2})]_{1}^{2}$.$A = 2 [(\frac{2}{2}\sqrt{4-4} + 2\sin^{-1}(1)) - (\frac{1}{2}\sqrt{4-1} + 2\sin^{-1}(\frac{1}{2}))]$.$A = 2 [(0 + 2(\frac{\pi}{2})) - (\frac{\sqrt{3}}{2} + 2(\frac{\pi}{6}))]$.$A = 2 [\pi - \frac{\sqrt{3}}{2} - \frac{\pi}{3}] = 2 [\frac{2\pi}{3} - \frac{\sqrt{3}}{2}] = \frac{4\pi}{3} - \sqrt{3}$.

The area of the smaller part between the circle $x^2+y^2=4$ and the line $x=1$ is . . . . . . sq. units.

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