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Question

* A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of (x+y)2 on the path AB traversed in a counter-clocwise sense is


A
π21
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B
π2+1
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C
π2
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D
1
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Solution

The correct option is B π2+1
Method I :
x2+y2 = 1
Let x = cos θ,
y = sinθ, θϵ[0,π2]
F = (x+y)2 and dr =rdθ
Now, cFdr=(x+y)2rdθ=π20(cosθ+sinθ)2dθ
=π20(1+sin2θ)dθ
=[θcos2θ2]π20=π2+1

Method II :
We have equation for path AB
x2+y2 = 1
Differentiating above equation w.r.t. y, we get
2xdxdy+2y=0

dxdy=yx
Let Φ=(x+y)2. Then
AB(x+y)2dr=AB(x+y)2(dx)2+(dy)2
=AB(x2+y2+2xy)1+(dxdy)2dy
=AB(x2+y2+2xy)(1+y2x2)dy
=AB(1+2xy).1x.dy
=AB(1x+2y)dy
=AB(11y2+2y)dy
=[sin1y+y2]10
=π2+1

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