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Math Help - HELP!! Volume of a solid w/ uniform cross sections of a parabola

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    HELP!! Volume of a solid w/ uniform cross sections of a parabola

    I didn't know how to explain the problem in words so I made a scan of it.




    Edit: Any help at all will be very appreciated.
    Last edited by sj9110; December 9th 2008 at 06:37 AM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sj9110 View Post
    I need to find the volume of a parabola a-x^2 where a is 0 to 10 and its also bounded by y=x or something. I don't know. Clearly I am very confused. I have a drawing of what the solid should look like.

    I didn't know how to explain the problem in words so I made a scan of it.




    Any help at all will be very appreciated.
    recall that the volume between x = a and x = b ( a \le b) is given by V = \int_a^b A(x)~dx, where A(x) is the formula for the cross-sectional area.

    in this case, our cross-sections are parabolas of the form y = a - x^2. but what is the area of this parabola that fits your constraints? recall that the area "under" the curve is given by the integral. so graph the curve y = a - x^2 for arbitrary a. find the intercepts and all that. now find the area under this curve. we see the area will be given by: A = \int_{\sqrt{a}}^{\sqrt{a}} (a - x^2)~dx = 2 \int_0^{\sqrt{a}}(a - x^2)~dx.

    Now, when you get the integral, replace the a with x, since in this case, a = x (the parabola is bounded by y = x, so the height is given by x, so that a = x since a is the height of our parabola above the x-axis). then go to the volume formula i gave you above to find the answer. x ranges from 0 up to 10
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    Solid with parabolic cross-section

    Hi -

    I attach a drawing of the problem as I understand it, with three axes: Ox, Oy and Oz.

    The values of x go from 0 to 10. At x = a, the parabola (which lies in a plane parallel to the y-z plane) has equation z=a-y^2. This parabola will have values of y from -\sqrt a to \sqrt a, and values of z from 0 to a.

    What you need to do, then, to find the volume of the solid enclosed by all these parabolas, the x-y plane and the plane x = 10, is:

    • Find the area A of the typical parabola shown. (Do this in the usual way with an integral, whose limits are y=-\sqrt a to y=\sqrt a.)
    • Replace a by x in your formula for A.
    • Now imagine x increasing by an amount \delta x. As it does so, the volume 'swept out' is approximately A\delta x. So the total volume will be \int_0^{10} A dx. So, replace A by your formula in terms of x, and then work out the integral.

    Have I given you enough to go on? Let me know if you want me to check your working.

    Grandad
    Attached Thumbnails Attached Thumbnails HELP!! Volume of a solid w/ uniform cross sections of a parabola-paraboliccrosssection.jpg  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Grandad View Post
    Hi -

    I attach a drawing of the problem as I understand it, with three axes: Ox, Oy and Oz.

    The values of x go from 0 to 10. At x = a, the parabola (which lies in a plane parallel to the y-z plane) has equation z=a-y^2. This parabola will have values of y from -\sqrt a to \sqrt a, and values of z from 0 to a.

    What you need to do, then, to find the volume of the solid enclosed by all these parabolas, the x-y plane and the plane x = 10, is:

    • Find the area A of the typical parabola shown. (Do this in the usual way with an integral, whose limits are y=-\sqrt a to y=\sqrt a.)
    • Replace a by x in your formula for A.
    • Now imagine x increasing by an amount \delta x. As it does so, the volume 'swept out' is approximately A\delta x. So the total volume will be \int_0^{10} A dx. So, replace A by your formula in terms of x, and then work out the integral.

    Have I given you enough to go on? Let me know if you want me to check your working.

    Grandad
    this is what i mean, i answered this problem in the other thread, giving the same solution (though i must admit, Grandad was a bit more elegant, using 3-dimensions and giving a nice diagram to boot--but it was still the same solution).
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