Anyone willing to verify my answer to a double integral? Thanks

- anonymous

Anyone willing to verify my answer to a double integral? Thanks

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- anonymous

what is the question?

- anonymous

\[\int\limits_{-3}^{2}\int\limits_{0}^{1}y^{2} x\]dy dx

- anonymous

And what was your answer?

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## More answers

- anonymous

Should I write it? Might it influence others' solution?

- anonymous

-5/6

- anonymous

You want us to verify your answer you should give your answer ;p

- anonymous

Lol, I meant see what answer you get and then I'll check if it's the same.

- anonymous

Oh. Yes I also got -5/6

- anonymous

Thanks!

- anonymous

No it is not -5/6

- anonymous

opps..ya it is, sorry

- anonymous

Yeah I got something different at first, but I thought it was dxdy, not dydx

- 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

If you ask, we may answer. If you don't ask we certainly will not help ;p

- anonymous

Haha, Ok here goes:

- anonymous

\[\int\limits_{-1}^{1}\int\limits_{0}^{\sqrt{1-x ^{2}}}(x ^{2}+y ^{2})^{3/2}\]dydx

- anonymous

Convert to polar.

- anonymous

\[\int\limits_{0}^{\pi}\int\limits_{0}^{1}r^3 r\ drd \Theta\]

- anonymous

Does that make sense? I'm getting Pi/5 for the answer.

- anonymous

Thanks dude, sorry about the late reply, my internet went and has finally JUST come back!

- anonymous

Can you help me a bit step by step... Thanks. How did you convert to polar?

- anonymous

is it the x=rcostheta and y=rsintheta?

- 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

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

So your radius goes from 0 to 1, and your angle goes from 0 to Pi.

- anonymous

Can you explain how you realised that it is the top half of a circle?

- 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

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

I understand the 0 to 1 limit of y. How did the 0 to Pi limit come about? Thanks

- 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

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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