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Math Help - Solid of Revolution

  1. #1
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    Solid of Revolution

    Im stuck on the following question as i dont understand the wording and what it is actually asking me to do.


    Consider the solid generated by revolving the region in the first quadrant between the lines y = x and y = x^2 about the y axis. The curve  y = x^2 lies on the outside of the solid.

    Find the volume of this region by dividing it up into small cylinders.

    any help would be most appreciated
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  2. #2
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    Shells (the little cylinders you mention):

    2{\pi}\int_{0}^{1}x(x-x^{2})dx

    washers:

    {\pi}\int_{0}^{1}(y-y^{2})dy
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  3. #3
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    thanks for the reply galactus but i dont quite understand what you mean
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  4. #4
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    Are you completely unfamiliar with volumes of revolution?.

    It's too much to explain it all from scratch.

    We are revolving the region between the two graphs around the y axis.

    This creates a solid with a volume. In this case, it looks kind of like a bowl.

    With shells, the cross sections are parallel to the axis around which we are

    revolving. therefore, we integrate w.r.t x. It is based on the area of a

    cylindrical shell. 2{\pi}\cdot(\text{average radius}) \cdot(\text{height})\cdot(\text{thickness})

    Which eventually gives us 2{\pi}\int_{a}^{b}xf(x)dx

    With washers, we have an infinite amount of circles we are finding the area

    of. {\pi}\int_{a}^{b}[f(x)]^{2}dx. See how it looks like the

    area of a circle?. {\pi}r^{2}

    This leads to the volume of said region.
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