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Thread: Region bounded by the graph....

  1. #1
    New Member
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    Region bounded by the graph....

    Let R be the region bounded by the y-axis and the graphs of y=((x^3)/(1+(x^2))) and y=4-2x.

    a.) Find the area of R
    -->What I think... is this where you do top-bottom or right-left (I never know which to use!!)

    b.) Find the volume of the solid generated when R is revolved about the x-axis.
    -->What I think... I have no idea on this one--or is this when you integrate?

    c.) The region R is the base of a solid. for this solid, each cross section perpendicular to the x-axis is a square. Find the volume of this solid.
    --> What I think... is this where you integrate??

    **Please help me get these started! I could use as much help as you can give ASAP! thanks sooooooooooo much!
    -Mooch

  2. #2
    Elite Member
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    I'll get you started with #1.

    The area between 2 continuous functions is [tex]\L\\\int_{a}^{b}[f(x)-g(x)]dx[/tex]

    Let f(x)=4-2x and g(x)=[tex]\L\\\frac{x^{3}}{1+x^{2}}[/tex]

    Set the two equations equal to each other and solve for x to find the limit(s) of integration. Try graphing the equations. You can see where they intersect and see what the limits of integration will be.

    The second problem is using 'disks' or 'washers'. This one is fairly strightforward. The volume of a solid of revolution using 'disks' is given by:

    [tex]\L\\{\pi}\int_{a}^{b}[f(x)]dx[/tex]. In your case:

    [tex]\L\\{\pi}\int_{a}^{b}[(f(x))^{2}-(g(x))^{2}]dx[/tex]


    For the 3rd problem, each cross-section has height f(x) and width dx. This would be a side of the square(if I am understanding the statement correctly).

    Now, how does that relate to the area of a square?.

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