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Multiple Integrals - Polar Coordinates

Is that sketch correct?

And I am asked to find M using suitable coordinates. Am I correct in thinking I need to change everything in polar form? (x^2 + y^2 = r^2)
IMAG0894[1].jpg164.1KB

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Original post by ChrisLorient
Is that sketch correct?

And I am asked to find M using suitable coordinates. Am I correct in thinking I need to change everything in polar form? (x^2 + y^2 = r^2)


Yes? No?
IMAG0895[1].jpg185.8KB
Original post by ChrisLorient
Is that sketch correct?

And I am asked to find M using suitable coordinates. Am I correct in thinking I need to change everything in polar form? (x^2 + y^2 = r^2)


Sketch looks fine.

And yes, your density function strongly suggests polar.

You should be able to write down xCx_C wth no working - considering odd/even functions.
Original post by ChrisLorient
Yes? No?


Yes.
Original post by ghostwalker
Sketch looks fine.

And yes, your density function strongly suggests polar.

You should be able to write down xCx_C wth no working - considering odd/even functions.


I have written down what I think Xc and Yc are, and so they are hopefully correct and ready for integrating when I got an answer for M.

I have worked M from the previous correct stage, but I am unsure as to whether I have done it correctly.
IMAG0896[1].jpg196.7KB
Original post by ChrisLorient
I have written down what I think Xc and Yc are, and so they are hopefully correct and ready for integrating when I got an answer for M.

I have worked M from the previous correct stage, but I am unsure as to whether I have done it correctly.


M looks fine.

For your xC,yCx_C,y_C, you seem to have inverted the ρ(x,y)\rho (x,y)

Note: xCx_C is doable in cartesian with no working - see previous post.
(edited 6 years ago)
Original post by ghostwalker
M looks fine.

For your xC,yCx_C,y_C, you seem to have inverted the ρ(x,y)\rho (x,y)

Note: xCx_C is doable in cartesian with no working - see previous post.


If as above, M is correct, how do I proceed with the integration? I am stuck at the stage shown in the last post.

And the Xc and Yc were given for me to use in the question. Apparently the moment over mass is the centre of mass for each coordinate, and thus the overall lamina.

I use x = rcostheta and y = rsintheta for the x and y in those equations?
(edited 6 years ago)
Original post by ChrisLorient
If as above, M is correct, how do I proceed with the integration? I am stuck at the stage shown in the last post.

"e-1" is just a constant.


And the Xc and Yc were given for me to use in the question. Apparently the moment over mass is the centre of mass for each coordinate, and thus the overall lamina.


That still doesn't explain why you have the e^r in the denominator, rather than the numerator. You had it correct when working out M.

E.g.

∫∫xρ(x,y)  dxdy\displaystyle\int\int x\rho(x,y)\;dxdy

Would become

∫∫rsin⁡θ×err×r  drdθ\displaystyle\int\int r\sin\theta \times \frac{e^r}{r}\times r \;drd\theta
Original post by ghostwalker
"e-1" is just a constant.



That still doesn't explain why you have the e^r in the denominator, rather than the numerator. You had it correct when working out M.

E.g.

∫∫xρ(x,y)  dxdy\displaystyle\int\int x\rho(x,y)\;dxdy

Would become

∫∫rsin⁡θ×err×r  drdθ\displaystyle\int\int r\sin\theta \times \frac{e^r}{r}\times r \;drd\theta


So integrating with respect to theta gives theta? And then subbing in pi for the upper limit would give pi?
(edited 6 years ago)
Original post by ChrisLorient
So integrating with respect to theta gives theta? And then subbing in pi for the upper limit would give pi?


I assume this is refering to M. You still have the constant there, other than that fine.
Original post by ghostwalker
I assume this is refering to M. You still have the constant there, other than that fine.


Starting from the integral that you said was correct for M, with working above to check if necessary.

I have tried, but no idea how to do it.
1508339073379-2113669251.jpg
Original post by ChrisLorient
Starting from the integral that you said was correct for M, with working above to check if necessary.

I have tried, but no idea how to do it.


∫0πe−1  dθ\displaystyle\int_0^\pi e-1\;d\theta

=[(e−1)θ]0π\displaystyle=\left[ ( e-1)\theta \right]_0^\pi

e-1 is just a constant like any other.
Original post by ghostwalker
∫0πe−1  dθ\displaystyle\int_0^\pi e-1\;d\theta

=[(e−1)θ]0π\displaystyle=\left[ ( e-1)\theta \right]_0^\pi

e-1 is just a constant like any other.


Oh, of course. e alone is a constant, so a constant minus a constant is a constant.

Do you simply get Pi(e-1) as the answer then?
(edited 6 years ago)
Original post by ChrisLorient
Oh, of course. e alone is a constant, so a constant minus a constant is a constant.

Do you simply get Pi(e-1) as the answer then?


Yes.
Original post by ghostwalker
Yes.


OK, so I have got Xc and Yc in polar coordinates. But what are their respective coordinates?
Original post by ChrisLorient
OK, so I have got Xc and Yc in polar coordinates. But what are their respective coordinates?


I can't make sense of that,
Original post by ghostwalker
I can't make sense of that,


I have these based on the equations for centre coordinates I was given.

15083985743791075324867.jpg

And the same but cos for the other.

But what are the limits? Is it 1 and 0 for x and Pi and 0 for y?
Original post by ChrisLorient
I have these based on the equations for centre coordinates I was given.


x is r cos theta, not r sin theta.

Other than that it's fine, and limits are the same as when you calculated M.

Why all the little mistakes - it's not the sort of thing I'd be expecting at this level?
Original post by ghostwalker
x is r cos theta, not r sin theta.

Other than that it's fine, and limits are the same as when you calculated M.

Why all the little mistakes - it's not the sort of thing I'd be expecting at this level?


Yeah, I noticed after taking the picture, but decided not to correct it and take the picture again.

So I now have them as follows.

1508417673698-1742458592.jpg

I will post again with my integration.
Original post by ChrisLorient
Yeah, I noticed after taking the picture, but decided not to correct it and take the picture again.

So I now have them as follows.

1508417673698-1742458592.jpg

I will post again with my integration.


This is my attempt so far. It all seems so wrong, as I have only one lamina problem under my belt, and it wasn't polar coordinates.

IMAG0914.jpg

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