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Math Help - Rotate About x-axis

  1. #1
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    Rotate About x-axis

    Find volume of solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.

    y = x^2, y = 4x-2x^2

    ABOUT X-AXIS.

    I set x^2 = 4x-2x^2 to find my limits of integration.
    I found my limits to be [0,2].
    See picture for more information.
    Attached Thumbnails Attached Thumbnails Rotate About x-axis-cam00368.jpg  
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  2. #2
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    Re: Rotate About x-axis

    Quote Originally Posted by nycmath View Post
    Find volume of solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.

    y = x^2, y = 4x-2x^2

    ABOUT X-AXIS.

    I set x^2 = 4x-2x^2 to find my limits of integration.
    I found my limits to be [0,2].
    See picture for more information.
    $x^2=4x - 2x^2$

    $3x^2 -4x =0$

    $3x\left(x-\frac 4 3 \right)=0$

    $x=0 \vee x=\frac 4 3$
    Thanks from nycmath
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    Re: Rotate About x-axis

    I made a mistake. The correct functions are
    y = x^2, y = 4x-x^2.
    Now, we can say that the limits of integration are [0,2].

    x^2 = 4x-x^2

    2x^2-4x = 0

    2x(x-2) = 0

    Solving for x, I get x = 0 and x = 2.
    How can the volume be 32pi/3?
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    Re: Rotate About x-axis

    Quote Originally Posted by nycmath View Post
    I made a mistake. The correct functions are
    y = x^2, y = 4x-x^2.
    Now, we can say that the limits of integration are [0,2].

    x^2 = 4x-x^2

    2x^2-4x = 0

    2x(x-2) = 0

    Solving for x, I get x = 0 and x = 2.
    How can the volume be 32pi/3?
    That's what the answer is. Just find

    $\pi \displaystyle{\int_0^2}\left(4x^2-x^2\right)^2-\left(x^2\right)^2~dx$

    do you understand this?
    Thanks from nycmath
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    Re: Rotate About x-axis

    Yes, it is the top function squared minus the bottom function squared.

    I can take it from here.

    Tell me, what if we are given y = sqrt(x),
    y = 0, x = 4 and told to rotate about the y-axis? What is the volume in this case?
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    Re: Rotate About x-axis

    Quote Originally Posted by nycmath View Post
    Yes, it is the top function squared minus the bottom function squared.

    I can take it from here.

    Tell me, what if we are given y = sqrt(x),
    y = 0, x = 4 and told to rotate about the y-axis? What is the volume in this case?
    $2\pi \displaystyle{\int_0^4}x \sqrt{x}~dx$

    $\sqrt{x}\geq 0~~\forall x$ so I can omit the absolute value sign.
    Thanks from nycmath
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  7. #7
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    Re: Rotate About x-axis

    Where did 2pi come from?
    I will check your reply in the morning.
    Good night and thank you.
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    Re: Rotate About x-axis

    Quote Originally Posted by nycmath View Post
    Where did 2pi come from?
    I will check your reply in the morning.
    Good night and thank you.
    Solid of revolution - Wikipedia, the free encyclopedia

    Look at the section on cylinder method.
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