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Math Help - The Jacobian

  1. #1
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    The Jacobian

    Here's my question:



    (a)

    \int^{2 \pi}_0 \int^1_0 [r^2(cos^2\theta + sin^2\theta)]rdr d\theta

    \int^{2 \pi}_0 \int^1_0 r^3 dr d \theta = \int^{2 \pi}_0 \frac{r^4}{a} |^1_0 d \theta

    =\frac{1}{4} \theta |^{2 \pi}_0 = \frac{\pi}{2}

    Is this correct?

    (b) So, is the Jacobian for this problem given by the following?

     <br />
J(x,y)= \frac{\partial(r,\theta)}{\partial (x,y)}=\begin{bmatrix} {\partial r\over \partial x} & {\partial r\over \partial y} \\ {\partial \theta\over \partial x} & {\partial \theta\over \partial y} \end{bmatrix}<br />

    If so, how can I obtain the four partials in the matrix?
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  2. #2
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    The answer for (a) looks correct to me.

    When you're changing to polar coordinates, you need to use the Jacobian. It is r, as you have used in (a).

    To calculate the Jacobian, you use the same idea as you have written, but the numerator and denominator should be switched. So

     <br />
\frac{\partial(x,y)}{\partial (r,\theta)}=\begin{bmatrix}  {\partial x\over \partial r} & {\partial x\over \partial \theta} \\  {\partial y\over \partial r} & {\partial y\over \partial \theta} \end{bmatrix}<br />

    Using x = r\cos\theta and y = r\sin\theta, the rest will follow.
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  3. #3
    Junior Member slovakiamaths's Avatar
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    first part of your problem is correct
    for secondone see attachment


    Attached Files Attached Files
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  4. #4
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    Quote Originally Posted by demode View Post
    Here's my question:



    (a)

    \int^{2 \pi}_0 \int^1_0 [r^2(cos^2\theta + sin^2\theta)]rdr d\theta

    \int^{2 \pi}_0 \int^1_0 r^3 dr d \theta = \int^{2 \pi}_0 \frac{r^4}{a} |^1_0 d \theta

    =\frac{1}{4} \theta |^{2 \pi}_0 = \frac{\pi}{2}

    Is this correct?

    (b) So, is the Jacobian for this problem given by the following?

     <br />
J(x,y)= \frac{\partial(r,\theta)}{\partial (x,y)}=\begin{bmatrix} {\partial r\over \partial x} & {\partial r\over \partial y} \\ {\partial \theta\over \partial x} & {\partial \theta\over \partial y} \end{bmatrix}<br />

    If so, how can I obtain the four partials in the matrix?
    The other responders have shown that. I want to point out that your use of " r drd\theta" already has the Jacobian in it! If u(x,y) and v(x,y) are a change of variable then dx dy= \frac{\partial (u, v)}{\partial (x, y)} dudv. Saying that dx dy= r dr d\theta is the same as saying \frac{\partial (r, \theta)}{\partial (x, y)}= r
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  5. #5
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    I followed Slovakiamaths' working and the Jacobian is J=r. So to answer part (b), the only point interior to R at which the Jacobian equals zero is (0,0), right?

    Furthermore, the question asks: "Prove that the mapping used is 1 to 1 on region R excluding the boundary points, and excluding the origin. Note: one way to do this is to produce the inverse mapping." I know that a mapping is one to one if distinct points in the r, \theta plane have distinct images in the xy-plane. But in this case how do I need to prove it?
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  6. #6
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    Okay, my mistake. Here's the mapping

    x = rcos\theta
    y = rsin\theta

    I don't know how to show it's one to one. I have to show that every point on the unit circle (except (1,0) ) is the mapping of exactly one angle in the interval 0 to 2π. But I don't know how to check that for all the points. Can anyone help?
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  7. #7
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    Are you aware that the sine and cosine are 1-1 on the interval [0,2pi)?
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  8. #8
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    Quote Originally Posted by maddas View Post
    Are you aware that the sine and cosine are 1-1 on the interval [0,2pi)?
    Yes, I think they are. But how do we show/prove it, in order to answer the question?
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