Question

evaluate the integral

Answer 1

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

Answer 2

dxdy

Answer 3

What have you tried so far?

Answer 4

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

Answer 5

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

Answer 6

no, seems hard to draw.

Answer 7

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

Answer 8

they are on the integral?

Answer 9

Yeah the are

Answer 10

There are 4 functions in all.

Answer 11

how would i draw it?

Answer 12

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

Answer 13

0<y<2 0<x<sqrt(2y-y^2)

Answer 14

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.

Answer 15

No

Answer 16

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

Answer 17

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

Answer 18

last one is a semi circle

Answer 19

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.

Answer 20

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?

Answer 21

yes i know of greens

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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.

Answer 28

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

Answer 29

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.

Answer 30

integral of 2sintheta=-2costheta

Answer 31

Yes correct, go on.

Answer 32

lol therefore this method wont work?

Answer 33

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

Answer 34

i c

Answer 35

:P

Answer 36

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

Answer 37

i gotta move onto other problems

Answer 38

Wait does this make 100% perfect sense?

\[-2 \cos \theta |_0^{\pi/2} = (-2 \cos (\pi/2)) - (-2 \cos (0)) = 0 + 2\]

Answer 39

Well good luck