TylerD
  • TylerD
evaluate the integral
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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TylerD
  • TylerD
\[\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{2y-y^2}}\frac{ 1 }{ \sqrt{x^2+y^2} }\]
TylerD
  • TylerD
dxdy
Kainui
  • Kainui
What have you tried so far?

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TylerD
  • TylerD
i was thinking of converting it to polar, but then wouldnt know what to do with the bounds
Kainui
  • Kainui
Have you tried drawing out the region you're integrating over?
TylerD
  • TylerD
no, seems hard to draw.
Kainui
  • Kainui
Well, I guess at the very least, what are the bounds of the region?
TylerD
  • TylerD
they are on the integral?
Kainui
  • Kainui
Yeah the are
Kainui
  • Kainui
There are 4 functions in all.
TylerD
  • TylerD
how would i draw it?
Kainui
  • Kainui
I'm not saying draw anything, just tell me what the 4 bounds are, they're equations.
TylerD
  • TylerD
0
Kainui
  • 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
  • Kainui
No
Kainui
  • Kainui
If you need a graphing calc to graph y=0 and y=2 then you need to slow down
Kainui
  • Kainui
You can use a graphing calc for the last one though ahha
TylerD
  • TylerD
last one is a semi circle
Kainui
  • 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
  • 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
  • TylerD
yes i know of greens
TylerD
  • TylerD
since we know the radius, im gonna try polar real quick
TylerD
  • TylerD
in the case of polar i just get pi/2, but the answer says its 2...
Kainui
  • 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
  • TylerD
i see, but that seems like something u just knew,
Kainui
  • Kainui
Well, I knew it cause I saw the picture and from the picture I recognized its graph in polar coordinates.
Kainui
  • 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
  • TylerD
only issue with that is, if u eval it u get -2
Kainui
  • 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
  • TylerD
integral of 2sintheta=-2costheta
Kainui
  • Kainui
Yes correct, go on.
TylerD
  • TylerD
lol therefore this method wont work?
Kainui
  • Kainui
This method only doesn't work if you believe cos(0)=0
TylerD
  • TylerD
i c
Kainui
  • Kainui
:P
TylerD
  • 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
  • TylerD
i gotta move onto other problems
Kainui
  • Kainui
Wait does this make 100% perfect sense? \[-2 \cos \theta |_0^{\pi/2} = (-2 \cos (\pi/2)) - (-2 \cos (0)) = 0 + 2\]
Kainui
  • Kainui
Well good luck

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