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Math Help - Volume of a Solid by Revolving

  1. #1
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    Question Volume of a Solid by Revolving



    The answers don't make sense with what I have learned.
    for A) about the x-axis shouldn't it be (Left - Right)

    Shouldn't it have been: PI (antiderivative) of [(0)^2 - (x^3)^2] dx

    Why did they put (1)^2 instead of 0 when 0 is on the left.

    I understand part B, but A doesn't make sense to me.
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  2. #2
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    The question stated y=1, which means it is capped off to y=1

    The integral is setup as \int_a^b \pi[f(x)^2-g(x)^2]dx

    Where f(x) is the graph "above" of g(x).

    1 is above the graph y=x^3 from the domain 0<x<1, therefore 1^2 is written first.
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  3. #3
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    I don't understand why it would be like that for \int_a^b \pi[f(x)-g(x)]dx

    Shouldn't it be Left - Right when it is like that?
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  4. #4
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    When doing it "about the X-axis" shouldn't it look like this
    \int_a^b \pi[Left-Right]dx

    Also why does the Y = 1 have any effect on that since it is dx and not dy?
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  5. #5
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    Quote Originally Posted by Brazuca View Post
    When doing it "about the X-axis" shouldn't it look like this
    \int_a^b \pi[Left-Right]dx

    Also why does the Y = 1 have any effect on that since it is dx and not dy?
    about the x-axis, the method of washers volume is

    V = \pi \int_a^b [f(x)]^2 - [g(x)]^2 \, dx

    where f(x) is the upper function and g(x) is the lower function.


    about the y-axis, the method of washers volume is

    V = \pi \int_c^d [f(y)]^2 - [g(y)]^2 \, dy

    where f(y) is the right function and g(y) is the left function.
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  6. #6
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    OHH That is so much easier than the way I learned it.
    Last edited by Brazuca; July 28th 2009 at 09:47 AM.
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