The answer is \( \dfrac1{\sqrt2} - \dfrac12\).
–
Pi Han Goh
·
4 months, 2 weeks ago

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@Pi Han Goh
–
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 \[ \\ x=g\left( u,v \right) \quad and\quad y=h\left( u,v \right) \].
–
Rishabh Deep Singh
·
4 months, 2 weeks ago

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@Rishabh Deep Singh
–
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?
–
Pi Han Goh
·
4 months, 2 weeks ago

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@Pi Han Goh
–
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.
–
Rishabh Deep Singh
·
4 months, 2 weeks ago

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@Rishabh Deep Singh
–
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?
–
Pi Han Goh
·
4 months, 2 weeks ago

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The answer is \( \dfrac1{\sqrt2} - \dfrac12\). – Pi Han Goh · 4 months, 2 weeks ago

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– Rishabh Deep Singh · 4 months, 2 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 \[ \\ x=g\left( u,v \right) \quad and\quad y=h\left( u,v \right) \].Log in to reply

– Pi Han Goh · 4 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?Log in to reply

– Rishabh Deep Singh · 4 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.Log in to reply

– Pi Han Goh · 4 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?Log in to reply