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Math Help - density

  1. #1
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    density

    1. Find the area, mass, and center of mass of a thin plate bounded by  x^2-y^2 =1 and  x = 4 with density  \rho(x,y) = x .

    Computing these, should be OK. Just wondering about the limits of integration. Would it be  x = 0 to  x = 4 and  y = 0 to  y = \sqrt{x^2-1} ?


    2. Let  f(x,y) = \begin{cases} 1, \ \ \ \ \ \text{if} \ x  \ \text{is irrational} \\ 4y^{3}, \ \ \text{if not} \end{cases}

    Show that  \int_{0}^{1} \left(\int_{0}^{1} f(x,y) \ dy \right) \ dx = 1 , but that  \int_{0}^{1} \left(\int_{0}^{1} f(x,y) \ dx \right) \ dy  does not exist.

    Then this implies that  \int_{0}^{1} f(x,y) \ dy = 1 or  \int f(x,y) \ dy = y . So we only integrate it when  f(x,y) = 1 ? And for the second case, the function may be similar to  \int_{0}^{1} \frac{1}{x^2} \ dx which does not exist?

    The only problem is, when we are integrating from  0 to  1 , we integrate over both rationals and irrationals. So maybe the second case deals with this, and that is why it doesn't exist. Whereas, in the first case, we keep  x constant, and integrate over the rationals between  0 and  1 ?
    Last edited by heathrowjohnny; February 23rd 2008 at 07:15 PM.
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    1. Find the area, mass, and center of mass of a thin plate bounded by  x^2-y^2 =1 and  x = 4 with density  \rho(x,y) = x .

    Computing these, should be OK. Just wondering about the limits of integration. Would it be  x = 0 to  x = 4 and  y = 0 to  y = \sqrt{x^2-1} ?
    [snip]
    No. Draw the graphs to see the region. Note that <br />
x^2-y^2 =1<br />
is a hyperbola with centre at (0, 0) and vertices at (1, 0) and (-1, 0) .....

    If first integrating with respect to y, the region is defined by -\sqrt{x^2-1} \leq y \leq \sqrt{x^2-1} and 1 \leq x \leq 4.

    Alternatively, if first integrating with respect to x, the region is defined by \sqrt{y^2+1} \leq x \leq 4 and -\sqrt{15} \leq y \leq \sqrt{15}.
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  3. #3
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    its not riemann integrable but it is lebesgue integral. so I dont think you can integrate first with  x (to be riemann integrable).
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    do you agree with me (i.e. its not riemann integrable but lebesgue integral)?
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  5. #5
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    Quote Originally Posted by heathrowjohnny View Post
    do you agree with me (i.e. its not riemann integrable but lebesgue integral)?
    You're referring to question 2 ......? I haven't had a close look at it yet, but my off the cuff answer is yes, I do agree with you (I may regret saying that ....) Someone like TPH would be better qualified than me to comment on on this question ......
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