evaluate the integral

- TylerD

evaluate the integral

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- TylerD

\[\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{2y-y^2}}\frac{ 1 }{ \sqrt{x^2+y^2} }\]

- TylerD

dxdy

- Kainui

What have you tried so far?

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## More answers

- TylerD

i was thinking of converting it to polar, but then wouldnt know what to do with the bounds

- Kainui

Have you tried drawing out the region you're integrating over?

- TylerD

no, seems hard to draw.

- Kainui

Well, I guess at the very least, what are the bounds of the region?

- TylerD

they are on the integral?

- Kainui

Yeah the are

- Kainui

There are 4 functions in all.

- TylerD

how would i draw it?

- Kainui

I'm not saying draw anything, just tell me what the 4 bounds are, they're equations.

- TylerD

0

- Kainui

Ahh ok I see. Yes, this is true, this is the overall region, however this is kind of confusing to write it this way if you just want to look at the boundary of the area, instead I suggest you look at it like this:
\[y=0\]\[y=2\]\[x=0\]\[x=\sqrt{2y-y^2}\]
Do your best to draw each of these all on the same graph, I expect it will be easy except for that last one so don't worry about that too much, but definitely try to figure out what that graph will look like.

- Kainui

No

- Kainui

If you need a graphing calc to graph y=0 and y=2 then you need to slow down

- Kainui

You can use a graphing calc for the last one though ahha

- TylerD

last one is a semi circle

- Kainui

That or you could see this:
\[x=\sqrt{2y-y^2}\]\[x^2 = 2y-y^2\]\[x^2+y^2-2y+1=1\]
\[x^2+(y-1)^2=1\]
This is a circle of radius 1 shifted up on the y-axis by 1.

- Kainui

Ok cool, so you have the region of integration, it's the algebra that made it complicated but it's really just half a circle so uhhh now what? I guess there are probably a couple ways to handle this integral now.
If you're having trouble coming up with ideas of what to do next, I'd probably try doing a substitution like \(y-1=u\) and leave x as it is, and then after that if it looks bad still, go to polar coordinates. That's just my first guess at it, try it out unless you wanna try some idea of your own.
Also, have you heard of Green's theorem? Or actually, are you expected to use it?

- TylerD

yes i know of greens

- TylerD

since we know the radius, im gonna try polar real quick

- TylerD

in the case of polar i just get pi/2, but the answer says its 2...

- Kainui

Ah ok you didn't adjust your bound correctly, specifically that hemispherical part looks like this:
https://www.desmos.com/calculator/dnvk7uivhy
So that's your new upper bound on r

- TylerD

i see, but that seems like something u just knew,

- Kainui

Well, I knew it cause I saw the picture and from the picture I recognized its graph in polar coordinates.

- Kainui

I don't know if its unreasonable to expect people to know what these graphs look like:
\[r=\sin \theta\] and \[r=\cos \theta\]
I'm not a teacher or anything I just sorta like doing calculus for fun ya know so I don't know if these are taught or what, but they're not like that crazy I think either.

- TylerD

only issue with that is, if u eval it u get -2

- Kainui

I can only say that I have struggled with integration for a long time but the thing that made it easier was to just take the pictures and geometrical shapes as the most important thing, and then their representation in coordinates seems to just become simple usually.
From our picture I can pretty quickly turn it into either integral:
\[\int_0^2 \int_0^{\sqrt{2y-y^2}} \frac{1}{\sqrt{x^2+y^2}} dx dy = \int_0^{\pi/2} \int_0^{2 \sin \theta} dr d \theta \]
Also I think you've made a mistake in evaluating it somewhere cause I get +2.

- TylerD

integral of 2sintheta=-2costheta

- Kainui

Yes correct, go on.

- TylerD

lol therefore this method wont work?

- Kainui

This method only doesn't work if you believe cos(0)=0

- TylerD

i c

- Kainui

:P

- TylerD

well, not sure if id be able to do a similar problem on my on for the final i have in 2 hours but

- TylerD

i gotta move onto other problems

- Kainui

Wait does this make 100% perfect sense?
\[-2 \cos \theta |_0^{\pi/2} = (-2 \cos (\pi/2)) - (-2 \cos (0)) = 0 + 2\]

- Kainui

Well good luck

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