I tried to put \[ x=rcos(\theta) \] and \[ y=rsin(\theta) \] then integration becomes \[ \iint _{ D }{ (r\sin { (\theta ))drd\theta } } \] but i dont know how to change limits of integration. Can you help me with how do i change the limits of integration if we put \[ \\ x=g\left( u,v \right) \quad and\quad y=h\left( u,v \right) \].

Instead of determining the limits of integration, why don't you evaluate the indefinite integral, \( \displaystyle \int \dfrac y{\sqrt{x^2+y^2}} \, dx \) as a start?

@Rishabh Deep Singh
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If you know the indefinite integral of \( \displaystyle \int \dfrac y{\sqrt{x^2+y^2}} \, dx \) already, why can't you evaluate \( \displaystyle \int_y^1 \dfrac y{\sqrt{x^2+y^2}} \, dx \) too?

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The answer is \( \dfrac1{\sqrt2} - \dfrac12\).

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I tried to put \[ x=rcos(\theta) \] and \[ y=rsin(\theta) \] then integration becomes \[ \iint _{ D }{ (r\sin { (\theta ))drd\theta } } \] but i dont know how to change limits of integration. Can you help me with how do i change the limits of integration if we put \[ \\ x=g\left( u,v \right) \quad and\quad y=h\left( u,v \right) \].

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Instead of determining the limits of integration, why don't you evaluate the indefinite integral, \( \displaystyle \int \dfrac y{\sqrt{x^2+y^2}} \, dx \) as a start?

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