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Average temperature of a hotplate

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data

    An 8-inch hot plate is described by the region R = {(x,y) | x2 + y2 16}. The temperature at the point (x,y) is given by T(x,y) = 400 cos(0.1 ), measured in degrees Fahrenheit. What is the average temperature of the hot plate?

    2. Relevant equations

    would i use the centers of mass equations and then plug in the x, y values that i get into the Temperature equation? (Can you even centers of mass for a temp average?)
    X= My / m and Y = Mx / m

    or should i set up a double integral or something? im confused!
     
  2. jcsd
  3. Apr 19, 2010 #2
    no. it has to do with the stationary solution of the heat transfer equation:


    [tex]
    \frac{\partial u}{\partial t} = a^{2} \nabla^{2} u,
    [/tex]

    which, under the assumption of stationarity [tex]{\partial u}/{\partial t} = 0[/tex], reduces to the Laplace equation:


    [tex]
    \nabla^{2} u = 0,
    [/tex]

    where u(x, y) is the stationary temperature distribution.

    There are powerful theorems for elliptic partial differential equations that enable you to find the solution.
     
  4. Apr 19, 2010 #3

    Dick

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    Homework Helper

    Oh, come on. It's not a PDE problem. Just integrate T(x,y) over the plate and divide by the area of the plate. As read, the average temperature is 400*cos(0.1), it's a constant. There's clearly some formatting problems with the original question. The region of the plate should be x^2+y^2<=16.
     
  5. Apr 19, 2010 #4
    I'm sorry. the original equation for T(x,y) = 400cos (0.1 sqrt(x^2 +y^2)) and yes the region is x^2 + y^2 <= 16. Very sorry about that..

    So, I was trying to use a double integral with polar coordinates..

    int (from 0 to 2pi) 2nd integral (from 0 to 4) [400cos( 0.1sqrt(r^r) )] *r dr d(theta)

    Is this the right way of going about it?
     
  6. Apr 19, 2010 #5

    Yes, and you should also divide by the total area of the region.
     
  7. Apr 19, 2010 #6
    Then How would i go about integrating 400cos(0.1*sqrt(r)) * r dr ??
     
  8. Apr 19, 2010 #7
    Make the substitution [tex]t = 0.1 \sqrt{r}[/tex]
     
  9. Apr 19, 2010 #8

    Dick

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    Science Advisor
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    400cos (0.1 sqrt(x^2 +y^2)) is 400*cos(0.1*r). There's no sqrt(r) here. Just integrate by parts.
     
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