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Question

Find the area bounded by the circle x2+y2=16 and the line √3y=x in the first quadrant, using integration.

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Solution

Equation of the circle is x2+y2=4 where radiusr=2unitsLet AB represent the line x=√3yPoint C is the intersectio of line and circle.Put x=√3y in x2+y2=4 we get3y2+y2=4 or 4y2=4 or y=±1When y=1⇒x=√3When y=−1⇒x=−√3As point C is in first quadrant C is (√3,1)Now,area of △OAC=Area△OCX+Area△XCA=∫√30y1dx+∫2√3y2dxPut x=√3y⇒y=x√3 So,y1=x√3x2+y2=4⇒y2=4−x2 or y2=√4−x2∴ Area of △OAC=∫√30x√3dx+∫2√3√4−x2dx .......(1)Let I1=∫√30x√3dx=1√3[x22]√30=12√3(3−0)=√32Now, I2=∫2√3√4−x2dx=[x2√4−x2+42sin−1x2]2√3 since ∫√a2−x2dx=12x√a2−x2+a22sin−1xa+c=12(2)√4−4−12√3√4−3+42sin−122−42sin−1√32=−√32+2[sin−1(1)−sin−1(√32)]=−√32+2[π2−π3]=−√32+π3Putting the value of I1 and I2 in (1)Area of △OAC=√32−−√32+π3=π3

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