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Multiple Integrals - Polar Coordinates Watch

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

    Edit: Refering to post #30
    In the third line of working, isn't it costheta and not - costheta ?

    Edit: Makes no difference to the final answer, but still.
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    (Original post by ChrisLorient)
    In the third line of working, isn't it costheta and not - costheta ?

    Edit: Makes no difference to the final answer, but still.
    Yep - should have been cos theta..
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    (Original post by ghostwalker)
    Yep - should have been cos theta..
    And when doing the integration, I feel I may have fudged it. I took r*e^r and used parts (like I did in the other question on multiple integrals), ignoring costheta.

    Is that wrong?
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    (Original post by ChrisLorient)
    And when doing the integration, I feel I may have fudged it. I took r*e^r and used parts (like I did in the other question on multiple integrals), ignoring costheta.

    Is that wrong?

    Since you're integrating with respect to r, then cos theta is just a multiplying constant for that part of the integration, So, as long as you didn't write it out in full without the theta, it's fine. You could in fact pull it out of the inner integral entirely.

    E.g. \displaystyle  \int_0^\pi \cos \theta \int_0^1 f(r) \;dr d\theta

    where f(r) is your function of r.
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    (Original post by ghostwalker)
    Since you're integrating with respect to r, then cos theta is just a multiplying constant for that part of the integration, So, as long as you didn't write it out in full without the theta, it's fine. You could in fact pull it out of the inner integral entirely.

    E.g. \displaystyle  \int_0^\pi \cos \theta \int_0^1 f(r) \;dr d\theta

    where f(r) is your function of r.
    So that is fine? Write out the thing, show that integration by parts gives 1, then carry on as I have?
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    Also, if density in this example is a function of x and y, is it a function of r and theta, or only theta? It becomes e^r/r, so I'm thinking only a function of r?

    So it would be p(r) = e^r/r and not p(r,theta) = e^r/r ?
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    (Original post by ChrisLorient)
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    So that is fine? Write out the thing, show that integration by parts gives 1, then carry on as I have?
    From the top line of your working downwards, looks fine. I don't have time to refresh my memory with the whole thread at the moment and deal with your other point - maybe at lunchtime.
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    (Original post by ChrisLorient)
    Also, if density in this example is a function of x and y, is it a function of r and theta, or only theta? It becomes e^r/r, so I'm thinking only a function of r?

    So it would be p(r) = e^r/r and not p(r,theta) = e^r/r ?
    To my mind, it's better to think of it as \rho(r,\theta) rather than \rho(r). That's because the problem is naturally two-dimensional so you need two variables to describe a point on the sheet e.g. someone would naturally ask you something like "what's the density at a distance 1.2 units from the origin, and pi/3 radians from the horizontal?" If you suppress one of them, you ignore or hide that fact.

    If you want to "see" the other variable, you could write, say:

    \rho(r,\theta)=f(\theta)\dfrac{e  ^r}{r} where f(\theta)=1

    Having said all that, it is very common to merely write \rho(r) and to let people infer from context that there is no angular dependency. so it doesn't matter much either way.
 
 
 
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