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Question

The value of 00ex2.ey2dxdy is

A
π2
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B
π
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C
π
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D
π2
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Solution

The correct option is D π2
Given integral is
I=00ex2.ey2dxdy
=00e(x2+y2)dxdy
Put x=rcosθ,y=rsinθ,|J|=r (Jacobian of polar coordinate system).
Then to cover whole 1st Quadrant θ varies from 0 to π2 and r varies from 0 to
So, I=00e(x2+y2).dxdy
=π/2θ=0r=0er2.(rdrdθ)
=π2πr=0r.er2.dr=π4 (Put r2=t)

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