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Find the area of the smaller part of the circle {x}^{{2}}+{y}^

Question

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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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}

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Solution: require Area=

\int y d x

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

\displaystyle\Rightarrow{x}={a}{\Sin{\theta}}

d x = a cos θ d θ

when

x = \frac{a}{2}, sin θ = \frac{1}{2}

θ = \frac{π}{4}

when

x = a, sin θ = 1

θ = \frac{π}{2}

= 2 \int_{\frac{π}{4}}^{\frac{π}{2}} (a cos θ) (a cos θ) d θ

= 2 a^{2} \int_{\frac{π}{4}}^{\frac{π}{2}} (cos 2 θ + 1) d θ

\displaystyle={a}^{{2}}{\left[{\int_{{\frac{\pi}{{2}}}}^{{\frac{\pi}{{4}}}}}{\Cos{{2}}}\theta{d}\theta+\int{1}{d}\theta\right]}

= a^{2} [[sin 2 \frac{θ}{2}] + [θ]]

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

= a^{2} [\frac{π}{4} - \frac{1}{2}]

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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	150