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Question Number 215390 by MathematicalUser2357 last updated on 05/Jan/25

∮_γ x^2 dx [γ: x^2 +y^2 =1]

∮γx2dx[γ:x2+y2=1]

Answered by MrGaster last updated on 05/Jan/25

solve:∫_0 ^(2π) (cos t)^2 (−sin t)dt[∵x=cos t,y=sin t,dx=−sin t dt] − ∫_(0 ) ^(2π) cos^2 t sin t dt =−[((cos^3 t)/3)]_0 ^(2π) =−(((cos^3 (2π))/3)−((cos^3 (0))/3)) =−((1/3)−(1/3)) =0

solve:∫02π(cost)2(−sint)dt[∵x=cost,y=sint,dx=−sintdt]−∫02πcos2tsintdt=−[cos3t3]02π=−(cos3(2π)3−cos3(0)3)=−(13−13)=0

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