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Math Help - Integral Applications

  1. #1
    Junior Member
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    Integral Applications

    Hey,

    I am trying to solve this question where I need to find the y-coordinate of the center of mass. Once again, I am having trouble dealing with the integral with the maximum function.

    I am assuming that to find the y coordinate, you multiply the function by y, since with only 2 integrals with two variables that is what we are supposed to do.

    Any suggestions on how to solve this problem, or mainly on how to eliminate the maximum would be great!

    Thanks
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  2. #2
    Super Member malaygoel's Avatar
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    y-co-ordinate of centre of mass=

    \frac{\int\int\int (y\rho) dxdydz}{\int\int\int(\rho) dxdydz}

    when integrating with respect to z, break the integral into two different integrals

    in one integral, set the limits of y from 0 to 1
    in other integral, sat the limit of y from 1 to 2.

    in the first of the above max(1,y^2) will be 1
    in the second of the above, it will be y^2
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  3. #3
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    Quote Originally Posted by malaygoel View Post
    y-co-ordinate of centre of mass=

    \frac{\int\int\int (y\rho) dxdydz}{\int\int\int(\rho) dxdydz}

    when integrating with respect to z, break the integral into two different integrals

    in one integral, set the limits of y from 0 to 1
    in other integral, sat the limit of y from 1 to 2.

    in the first of the above max(1,y^2) will be 1
    in the second of the above, it will be y^2
    That makes sense. So would it basically be like a preset condition like for all y such that 0 < y < 1...the integration would look one way and then another way for greater y values?

    Therefore, when you do the final integrate with respect to y, for y = 0 you would sub it into the first integrated equation, and for y = 2 you would sub into the second integrated equation..

    If I misunderstood you, please let me know...and thanks again for your help!
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  4. #4
    Super Member malaygoel's Avatar
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    Evaluate:
    <br />
\frac{\int_1^2\int_{y^2}^4 \int_{-y}^{\frac{-yz}{y+z}} (y\rho) dxdzdy+\int_0^1\int_{1}^4 \int_{-y}^{\frac{-yz}{y+z}} (y\rho) dxdzdy}{\int_1^2\int_{y^2}^4 \int_{-y}^{\frac{-yz}{y+z}} (\rho) dxdzdy+\int_0^1\int_{1}^4 \int_{-y}^{\frac{-yz}{y+z}} (\rho) dxdzdy}<br />
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