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Explain how a multiple integral may be expressed in a different coordinate system

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    The two halves of this question seem oddly unrelated. Assuming that I'm right in thinking that the second part is nothing to do with the first, could someone help me work out how to answer the second part?

    I know that the volume element, dxdydz, becomes the modulus of the jacobian times dudvdw, but what happens to the region, R? Do you just convert the limits for x, y and z into terms of u, v and w? I'm finding changing limits here very confusing.
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    Can anyone help?
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    I can't see much relation between the first and second parts of this question (but I haven't actually tried doing the question). As for the second part, you seem to be along the right lines. The region R doesn't change when you change coordinates; all that changes is the way you express it. If u,v,w is a change of coordinates, then this means that u,v,w are all bijective functions of x,y,z. In particular, they have inverses, so x,y,z are bijective functions of u,v,w, and so the cone is parametrised by z(u,v,w)^2 = x(u,v,w)^2 + y(u,v,w)^2. Then the limits of the integral will be precisely the set of (u,v,w) for which this equation is satisfied.
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    (Original post by nuodai)
    I can't see much relation between the first and second parts of this question (but I haven't actually tried doing the question). As for the second part, you seem to be along the right lines. The region R doesn't change when you change coordinates; all that changes is the way you express it. If u,v,w is a change of coordinates, then this means that u,v,w are all bijective functions of x,y,z. In particular, they have inverses, so x,y,z are bijective functions of u,v,w, and so the cone is parametrised by z(u,v,w)^2 = x(u,v,w)^2 + y(u,v,w)^2. Then the limits of the integral will be precisely the set of (u,v,w) for which this equation is satisfied.
    I think that makes sense to me, but I don't understand how it's possible to get the new limits in a useable form to finish the question without knowing what the functions relating u, v and w to x, y and z are - because z(u,v,w)^2 = x(u,v,w)^2 + y(u,v,w)^2 doesn't seem to have a solution. :confused:
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    (Original post by 99wattr89)
    I think that makes sense to me, but I don't understand how it's possible to get the new limits in a useable form to finish the question without knowing what the functions relating u, v and w to x, y and z are - because z(u,v,w)^2 = x(u,v,w)^2 + y(u,v,w)^2 doesn't seem to have a solution. :confused:
    How could you possibly solve the equation without knowing what u,v,w are? If you know what they are then you might have a hope of solving it. So by saying "solve z^2=x^2+y^2 for u,v,w", you've said as much as you can until you know what u,v,w are. But the question is asking a general question, so you're not expected to give an explicit form of an answer (not least because such an expression doesn't exist); the specifics come later on.

    For instance, we could use cylindrical polar coordinates (u,v,w)=(r, \varphi, z), so that we have x = r \cos \varphi, y = r \sin \varphi and z = z. Then the equation of the surface, z^2 = x^2 + y^2, becomes z^2 = r^2. Note that this doesn't depend on \varphi, so \varphi can take any value and as long as r=\left|z\right| the equation is still satisfied. (This makes sense since all the equation above tells you is that for a given z, the cross-section of the graph through that point on the z-axis is just a circle of radius |z|, which is why it looks like a cone [or rather two cones stuck together at their tips].) So we can integrate over \theta, z by taking 1 \le z \le 2 (which is the given limit) and 0 \le \theta < 2\pi, and setting r=z in the integral and multiplying by the Jacobian.

    This is just one example of a parametrization. (It just happens to be probably the easiest parametrization to use in this case.)
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    (Original post by nuodai)
    How could you possibly solve the equation without knowing what u,v,w are? If you know what they are then you might have a hope of solving it. So by saying "solve z^2=x^2+y^2 for u,v,w", you've said as much as you can until you know what u,v,w are. But the question is asking a general question, so you're not expected to give an explicit form of an answer (not least because such an expression doesn't exist); the specifics come later on.

    For instance, we could use cylindrical polar coordinates (u,v,w)=(r, \varphi, z), so that we have x = r \cos \varphi, y = r \sin \varphi and z = z. Then the equation of the surface, z^2 = x^2 + y^2, becomes z^2 = r^2. Note that this doesn't depend on \varphi, so \varphi can take any value and as long as r=\left|z\right| the equation is still satisfied. (This makes sense since all the equation above tells you is that for a given z, the cross-section of the graph through that point on the z-axis is just a circle of radius |z|, which is why it looks like a cone [or rather two cones stuck together at their tips].) So we can integrate over \theta, z by taking 1 \le z \le 2 (which is the given limit) and 0 \le \theta < 2\pi, and setting r=z in the integral and multiplying by the Jacobian.

    This is just one example of a parametrization. (It just happens to be probably the easiest parametrization to use in this case.)
    I think I understand what you're saying. Thank you very much for your help.

    The answer I got for the final part seems a bit funny, PiLn2, but I think the method was right, and that's what matters.
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    (Original post by 99wattr89)
    I think I understand what you're saying. Thank you very much for your help.

    The answer I got for the final part seems a bit funny, PiLn2, but I think the method was right, and that's what matters.
    That's the right answer, so you probably did something right :p:
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    (Original post by nuodai)
    That's the right answer, so you probably did something right :p:
    Even better! Thanks for all your help!

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