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Find the area of the smaller part of the circle x^2 + y^2 = a^

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Find the area of the smaller part of the circle $x^{2} + y^{2} = a^{2}$ cut off by the line $x = \frac{a}{2}$ .

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Amelia Discussed

Find the area of the smaller part of the circle

x^{2} + y^{2} = a^{2}

cut off by the line

x = \frac{a}{2}

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

The area of the smaller part of the circle,

x^{2} + y^{2} = a^{2}

, cut off by the line,

x = \frac{a}{2}

, is the are ABCDA.
It can be observed that the area ABCD is symmetrical about x-axis.
\therefore \text{Area} \ ABCD =2\times \mbox{Area} \ ABC

Area of

A BC

= \int_{\frac{a}{2}}^{a} y d x

= \int_{\frac{a}{2}}^{a} a^{2} - x^{2} d x

= [\frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}]_{\frac{a}{2}}^{a}

= [\frac{a ^{2}}{2} (\frac{π}{2}) - \frac{a}{2 2} a^{2} - \frac{a ^{2}}{2} - \frac{a ^{2}}{2} sin^{- 1} (\frac{1}{2})]

= \frac{a ^{2} π}{4} - \frac{a}{2 2} \cdot \frac{a}{2} - \frac{a ^{2}}{2} (\frac{π}{4})

= \frac{a ^{2} π}{4} - \frac{a ^{2}}{4} - \frac{a ^{2} π}{8}

= \frac{a ^{2}}{4} [π - 1 - \frac{π}{2}]

= \frac{a ^{2}}{4} [\frac{π}{2} - 1]

\Rightarrow \mbox{ Area} \ ABCD =2\left [\displaystyle \dfrac {a^2}{4}\left (\frac {\pi}{2}-1\right )\right ]=\dfrac {a^2}{2}\left (\dfrac {\pi}{2}-1\right )

Therefore, the area of smaller part of the circle,

x^{2} + y^{2} = a^{2}

, cut off by the line,

x = \frac{a}{2}

, is

\frac{a ^{2}}{2} (\frac{π}{2} - 1)

sq.units.

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Question Text	Find the area of the smaller part of the circle $x^{2} + y^{2} = a^{2}$ cut off by the line $x = \frac{a}{2}$ .
Answer Type	Text solution:1
Upvotes	151

Find the area of the smaller part of the circle x2+y2=a2 cut off by the line x=2​a​.

Text solutionVerified

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Find the area of the smaller part of the circle $x^{2} + y^{2} = a^{2}$ cut off by the line $x = \frac{a}{2}$ .