Double integration

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

#Calculus

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

What have you tried? Where are you stuck on?

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

Pi Han Goh - 4 years, 2 months 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 $\\ x=g\left( u,v \right) \quad and\quad y=h\left( u,v \right)$ .

Rishabh Deep Singh - 4 years, 2 months 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?

Pi Han Goh - 4 years, 2 months ago

@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 years, 2 months ago

@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 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$