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# Multiple Integrals - Polar Coordinates watch

1. (Original post by ChrisLorient)
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.

Attachment 697056
Looks fine.
2. (Original post by ghostwalker)
Looks fine.

And usually with x-y coordinates I can get a feel for where it lies. But I can't tell where this would lie on the semi-circle we have.

Edit: Gives 0, which is what we would expect given symmetry, no? Although that doesn't mean it is correct, as many ways could give 0.
3. (Original post by ChrisLorient)

And usually with x-y coordinates I can get a feel for where it lies. But I can't tell where this would lie on the semi-circle we have.

Edit: Gives 0, which is what we would expect given symmetry, no? Although that doesn't mean it is correct, as many ways could give 0.
Fine.

Since x_C is zero you know what line it lies on. You just need to work out y_C now.
4. (Original post by ghostwalker)
Fine.

Since x_C is zero you know what line it lies on. You just need to work out y_C now.
Yc = (2e) / (2*Pi (e-1))

This gives me 0.504 (3 d.p.)

Sorry for no working, but my phone is playing up. Shall attach when I can.

Does that seem as though it is reasonable though?
5. (Original post by ghostwalker)
Fine.

Since x_C is zero you know what line it lies on. You just need to work out y_C now.
Actually, isn't esin(Pi) and esin(0) both 0, and so it's the fact it's something times 0 which makes it zero, meaning my numerator for Xc is incorrect.

Again, sorry for lack of image. But it's the one three posts up from this.
6. (Original post by ChrisLorient)
Actually, isn't esin(Pi) and esin(0) both 0, and so it's the fact it's something times 0 which makes it zero, meaning my numerator for Xc is incorrect.

Again, sorry for lack of image. But it's the one three posts up from this.
Yes, they're both zero.

I assumed you'd evaluated sin(0) as 0, and were just left with e*sin(pi), so it looked correct.
7. (Original post by ChrisLorient)
...
Sorry.

I just realised there is an error in the working for x_C, and I presume there will be one as well in y_C

You'll need to do something similar to what you did with the M working, because of the re^r term.
8. (Original post by ghostwalker)
Sorry.

I just realised there is an error in the working for x_C, and I presume there will be one as well in y_C

You'll need to do something similar to what you did with the M working, because of the re^r term.
Integrating with respect to r:

e^r (r-1)

So:

e^r (r-1) * cos(theta) dtheta

Is that correct?

Edit:

That is the overall problem. And my question is whether there is a mistake in Xc and Yc, as I have seen it written where X uses y and Y uses X (for the moment). Just wanting to make sure the question has no error, and to also help you if my attachments are making you lose track.

Attachment 697166
9. (Original post by ChrisLorient)
Integrating with respect to r:

e^r (r-1)

So:

e^r (r-1) * cos(theta) dtheta

Is that correct?
Yes.

Edit:

That is the overall problem. And my question is whether there is a mistake in Xc and Yc, as I have seen it written where X uses y and Y uses X (for the moment). Just wanting to make sure the question has no error, and to also help you if my attachments are making you lose track.

Attachment 697166
The form for x_C and y_C given in the question are the ones I would expect.

For working out x_C, the x-coordinate of the centre of mass, we use x, with the density. Can't see it being done any other way. If you have an example where that's not the case, I suggest posting a link/photo.
10. (Original post by ghostwalker)
Yes.
11. (Original post by ghostwalker)
Yes.

The form for x_C and y_C given in the question are the ones I would expect.

For working out x_C, the x-coordinate of the centre of mass, we use x, with the density. Can't see it being done any other way. If you have an example where that's not the case, I suggest posting a link/photo.
And for Yc.

12. (Original post by ChrisLorient)
Don't understand what you're asking. Can't see anything wrong with what you've written there.

Edit: Refering to post #30
13. (Original post by ghostwalker)
Don't understand what you're asking. Can't see anything wrong with what you've written there.
Now I'm questioning my own working.

Should I have r^2 e^r cos(theta) dr dtheta ?

Does it look as though I have forgot the r when changing dx dy to polar for X and Y?

I know I did it correctly when calculating M, but I think I may have missed it.
14. (Original post by ChrisLorient)
Now I'm questioning my own working.

Should I have r^2 e^r cos(theta) dr dtheta ?

Does it look as though I have forgot the r when changing dx dy to polar for X and Y?

I know I did it correctly when calculating M, but I think I may have missed it.
There was a 1/r from the density function that cancels it.
15. (Original post by DFranklin)
There was a 1/r from the density function that cancels it.
You think I have it correct then?

My only concern is the value is 0.185 for Yc (assuming the denominator does indeed 'flip' when multiplied by -1), and for some reason that seems wrong.
16. (Original post by ChrisLorient)
And for Yc.

From post #31

When you integrate with respect to r, and evaluate the two end points, you have correctly the upper end as 0 and the lower end as -1.

Then (0) - (-1) = 1, so there is no minus sign associated with the sin theta in the third line of working.

Edit: And subsequently you will have minus cos, and with evaluation at the end points you should have -[ (-1) - (1) ]= 2
17. (Original post by ChrisLorient)
You think I have it correct then?

My only concern is the value is 0.185 for Yc (assuming the denominator does indeed 'flip' when multiplied by -1), and for some reason that seems wrong.
See my previous post - including the edit.
18. (Original post by ghostwalker)
From post #31

When you integrate with respect to r, and evaluate the two end points, you have correctly the upper end as 0 and the lower end as -1.

Then (0) - (-1) = 1, so there is no minus sign associated with the sin theta in the third line of working.

Edit: And subsequently you will have minus cos, and with evaluation at the end points you should have -[ (-1) - (1) ]= 2
Sin then integrates to -cos, which gives 1 and -1 using the limits. That gives (1)-(-1), which is 2.

So the answer is twice the previous value, and a much more reasonable number at 0.370

Edit: Beat me to it. No surprise there.
19. (Original post by ChrisLorient)
Sin then integrates to -cos, which gives 1 and -1 using the limits. That gives (1)-(-1), which is 2.

So the answer is twice the previous value, and a much more reasonable number at 0.370

Edit: Beat me to it. No surprise there.
And breathe.
20. (Original post by ghostwalker)
And breathe.
Thanks so much for the help, and especially the patience. I made a meal of this question, I know.

PRSOM.

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