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# Double integration

$\int_0^1 \int_y^1 \dfrac{y \, dx \; dy}{\sqrt{x^2+y^2}} = \, ?$

Note by Rishabh Deep Singh
10 months, 3 weeks ago

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What have you tried? Where are you stuck on?

The answer is $$\dfrac1{\sqrt2} - \dfrac12$$.

- 10 months, 3 weeks ago

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 $\\ x=g\left( u,v \right) \quad and\quad y=h\left( u,v \right)$.

- 10 months, 2 weeks ago

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?

- 10 months, 2 weeks ago

I know that your integral will of the form arcsinh(x) but i want to learn how do i change the limits of the integration.

- 10 months, 2 weeks ago

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?

- 10 months, 2 weeks ago