Double integral with polar coordinates

[Archimedess]

Homework Statement

Let $A=\{(x,y)\in\mathbb{R}^2| 1\leq y \leq 2, x\geq 0, x^2+(y-1)^2\leq1\}$ then $\iint_A\frac{y}{x^2+y^2}$ is

Relevant Equations

$x=r\cos\theta$
$y=r\sin\theta$

Hello there,
I'm struggling in this problem because i think i can't find the right $\theta$ or $r$

Here's my work:

$\pi/4\leq\theta\leq\pi/2$
and
$0\leq r\leq 2\sin\theta$

So the integral would be: $\int_{\pi/4}^{\pi/2}\int_{0}^{2\sin\theta}\sin\theta dr d\theta$

Which is equal to: $\pi/4+1/2$ but this is not the right solution..($1/2$ is the correct one)

Any help? Thank you in advance!

 

[PeroK]

Why are you using polar coordinates?

 

[Archimedess]

I don't know another way to solve it.. i mean is not required but is the only way i know this time

 

[PeroK]

I don't know another way to solve it.. i mean is not required but is the only way i know this time

Have you drawn a diagram of the region $A$?

 

[Archimedess]

Have you drawn a diagram of the region $A$?

Yes, i notice that $1\leq y \leq 2$ and $0\leq x \leq ?$ the half semicircle

 

[PeroK]

Yes, i notice that $0\leq x \leq 1$ and $1\leq y \leq ?$ the half semicircle

Your original post says $1 \le y \le 2$.

 

[Archimedess]

Your original post says $1 \le y \le 2$.

Yes, sorry edited my post

The problem is now how do i write that half semicircle

 

[PeroK]

Yes, sorry edited my post

The problem is now how do i write that half semicircle

Is it $0 \le y \le 2$?

 

[Archimedess]

You'll need to change coordinates first before you can easily use polar coordinates.

Sure, that's what i did in my post i think..

I was trying something different like: $\int_{1}^{2}\int_{0}^{\sqrt{1-(y-1)^2}}\frac{y}{x^2+y^2}dxdy$

But i think is way more easy with polar coordinates once you find the right values..

 

[Archimedess]

Is it $0 \le y \le 2$?

Nope.. $1 \le y \le 2$ is correct

 

[PeroK]

Sure, that's what i did in my post i think..

I was trying something different like: $\int_{1}^{2}\int_{0}^{\sqrt{1-(y-1)^2}}\frac{y}{x^2+y^2}dxdy$

But i think is way more easy with polar coordinates once you find the right values..

You should be able to do the $dy$ integral first. But, it gets messy after that.

 

[PeroK]

Sure, that's what i did in my post i think..

I was trying something different like: $\int_{1}^{2}\int_{0}^{\sqrt{1-(y-1)^2}}\frac{y}{x^2+y^2}dxdy$

But i think is way more easy with polar coordinates once you find the right values..

You're right. It does come out more easily using polar coodinates. But,

Here's my work:

$\pi/4\leq\theta\leq\pi/2$
and
$0\leq r\leq 2\sin\theta$

This is not right. The lower limit for $r$ is not $0$.

 

[Archimedess]

This is not right. The lower limit for $r$ is not $0$.
[/QUOTE]

Should it be 1? Even if i use 1 the solution is not right

 

[PeroK]

Should it be 1? Even if i use 1 the solution is not right

It's a function of $\theta$. It can't be constant.