anonymous
  • anonymous
Anyone willing to verify my answer to a double integral? Thanks
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
what is the question?
anonymous
  • anonymous
\[\int\limits_{-3}^{2}\int\limits_{0}^{1}y^{2} x\]dy dx
anonymous
  • anonymous
And what was your answer?

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anonymous
  • anonymous
Should I write it? Might it influence others' solution?
anonymous
  • anonymous
-5/6
anonymous
  • anonymous
You want us to verify your answer you should give your answer ;p
anonymous
  • anonymous
Lol, I meant see what answer you get and then I'll check if it's the same.
anonymous
  • anonymous
Oh. Yes I also got -5/6
anonymous
  • anonymous
Thanks!
anonymous
  • anonymous
No it is not -5/6
anonymous
  • anonymous
opps..ya it is, sorry
anonymous
  • anonymous
Yeah I got something different at first, but I thought it was dxdy, not dydx
anonymous
  • anonymous
:)
anonymous
  • anonymous
lol! Thanks! Just making sure I know what I'm doing... I've got another similar question which I haven't answered yet as I need help answering it. Want to help out?
anonymous
  • anonymous
If you ask, we may answer. If you don't ask we certainly will not help ;p
anonymous
  • anonymous
Haha, Ok here goes:
anonymous
  • anonymous
\[\int\limits_{-1}^{1}\int\limits_{0}^{\sqrt{1-x ^{2}}}(x ^{2}+y ^{2})^{3/2}\]dydx
anonymous
  • anonymous
Convert to polar.
anonymous
  • anonymous
\[\int\limits_{0}^{\pi}\int\limits_{0}^{1}r^3 r\ drd \Theta\]
anonymous
  • anonymous
Does that make sense? I'm getting Pi/5 for the answer.
anonymous
  • anonymous
Thanks dude, sorry about the late reply, my internet went and has finally JUST come back!
anonymous
  • anonymous
Can you help me a bit step by step... Thanks. How did you convert to polar?
anonymous
  • anonymous
is it the x=rcostheta and y=rsintheta?
anonymous
  • anonymous
\[r^2 = x^2 + y^2 \rightarrow r = (x^2 + y^2)^{1/2} \rightarrow (x^2 + y^2)^{3/2} = r^3\] \[dydx = r\ drd \Theta\]
anonymous
  • anonymous
Then you have to convert your limits. Since x goes from -1 to 1 and y goes from 0 to sqrt(1+x^2) you can see that you're integrating over the top half of a circle with radius 1.
anonymous
  • anonymous
So your radius goes from 0 to 1, and your angle goes from 0 to Pi.
anonymous
  • anonymous
Can you explain how you realised that it is the top half of a circle?
anonymous
  • anonymous
Because the y values go from 0 to sqrt(1+x^2) Those are all positive y values inside the circle x^2 + y^2 = 1 If it had said -sqrt(1+x^2) then it would have been the bottom half of the circle.
anonymous
  • anonymous
If you graph the line y = sqrt(1+x^2) you'll see it forms the top half of a circle. Integrating from y = 0 to y=sqrt(1+x^2) will be integrating over that same half-circle.
anonymous
  • anonymous
I understand the 0 to 1 limit of y. How did the 0 to Pi limit come about? Thanks
anonymous
  • anonymous
You have to look at both pieces together. If y goes from 0 to the top of the half circle and x goes from -1 to 1 then the area you're covering is the top half of the circle. That means your r will go from 0 to 1 and your Theta will go from 0 to pi. cos(0) = 1 = x, cos(pi) = -1 = x.
anonymous
  • anonymous
Oh! Sorry about that! Theta goes from 0 to Pi, I was thinking about x... Understand that part now! Thanks a lot for your help!

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