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Math Help - Integration c of mass

  1. #1
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    Integration c of mass

    Hi, Can some one Please help. centre mas question
    The 'question' is in blue on my drawing, I have attempted it but I dont believe it is correct ,I made up my own value for variable c , as they didnt give values
    (I dont really understand it)
    Attached Thumbnails Attached Thumbnails Integration c of mass-integr-prob.jpg  
    Last edited by wolfhound; February 20th 2010 at 12:07 PM.
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  2. #2
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    Quote Originally Posted by wolfhound View Post
    ,I made up my own value for variable c , as they didnt give values
    For definite integrals you can ignore the constant term c as it will cancel out after inputing the terminals.
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    \bar{y} = \frac{\int_{-c}^c x^4 \, dx}{2\int_{-c}^c x^2 \, dx}<br />

    taking advantage of symmetry ...

    \bar{y} = \frac{\int_0^c x^4 \, dx}{2\int_0^c x^2 \, dx}<br />

    \bar{y} = \frac{\frac{c^5}{5}}{\frac{2c^3}{3}} = \frac{3c^2}{10}<br />

    so, if c = 3 , then the y-coordinate of the region's centroid is \bar{y}= \frac{27}{10} , and the x-coordinate is \bar{x} = 0 as evidenced by the function's symmetry to the y-axis.
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    Quote Originally Posted by skeeter View Post
    \bar{y} = \frac{\int_{-c}^c x^4 \, dx}{2\int_{-c}^c x^2 \, dx}<br />

    taking advantage of symmetry ...

    \bar{y} = \frac{\int_0^c x^4 \, dx}{2\int_0^c x^2 \, dx}<br />

    \bar{y} = \frac{\frac{c^5}{5}}{\frac{2c^3}{3}} = \frac{3c^2}{10}<br />

    so, if c = 3 , then the y-coordinate of the region's centroid is \bar{y}= \frac{27}{10} , and the x-coordinate is \bar{x} = 0 as evidenced by the function's symmetry to the y-axis.
    Thanks!
    I am just a bit puzzled why do they say x= -c and x=c and c is positive,
    so if I put the value of c=3 why dont I need to give -c a value ?or how does -c become 0?
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  5. #5
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    If f(x) is even, then \int_{-a}^a f(x) dx = 2 \int_{0}^a f(x) dx.

    Just made your integral a little bit less annoying.
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    Quote Originally Posted by o_O View Post
    If f(x) is even, then \int_{-a}^a f(x) dx = 2 \int_{0}^a f(x) dx.

    Just made your integral a little bit less annoying.
    Thank you but I dont understand,
    Could you explain this please:I am just a bit puzzled why do they say x= -c and x=c and c is positive,
    so if I put the value of c=3 why dont I need to give -c a value ?or how does -c become 0?
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  7. #7
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    They are saying that your region is bounded by the parabola y=x^2 and the two vertical lines x = -c and x = c where c is just some positive constant (the c>0 part).

    So when you set up your integral to find \bar{y}, we have: \bar{y} = \frac{\int_{-c}^c x^4 \, dx}{2\int_{-c}^c x^2 \, dx}

    Now, if c=3, then we would have: \bar{y} = \frac{\int_{-3}^3 x^4 \, dx}{2\int_{-3}^3 x^2 \, dx}

    You can solve that directly if you want. But what we noted earlier is that both y= x^4 and y = x^2 are even functions, and using the fact I gave you in my last post, we can convert each integral so that we have:

    \bar{y} = \frac{\int_{-3}^3 x^4 \, dx}{2\int_{-3}^3 x^2 \, dx} = \frac{{\color{red}2}\int_{{\color{red}0}}^3 x^4 \, dx}{2 \cdot {\color{red}2}\int_{{\color{red}0}}^3 x^2 \, dx} =  \frac{\int_{0}^3 x^4 \, dx}{2\int_{0}^3 x^2 \, dx}

    Doesn't really do anything too much but it makes evaluating the definite integral a little easier. We like zeroes .. kind of..
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  8. #8
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    Quote Originally Posted by wolfhound View Post
    Thank you but I dont understand,
    Could you explain this please:I am just a bit puzzled why do they say x= -c and x=c and c is positive,
    so if I put the value of c=3 why dont I need to give -c a value ?or how does -c become 0?
    -c did not become zero.

    Both functions, x^4 and x^2 are symmetrical to the y-axis ... as such, I took advantage of the symmetry by writing \int_{-c}^c x^4 \, dx = 2\int_0^c x^4 \, dx and \int_{-c}^c x^2 \, dx = 2\int_0^c x^2 \, dx .
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  9. #9
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    Smile

    I see thanks!
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