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Math Help - Volume for region rotated around the x-axis

  1. #1
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    Volume for region rotated around the x-axis

    I just need help setting them up, these are different because they have two y =. What formula would you use? thanks

    5) y
    = 2x, y = 2, x = 0



    6) y
    = x2 + 5, y = 2x + 5
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  2. #2
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    Quote Originally Posted by Collegeboy110 View Post
    I just need help setting them up, these are different because they have two y =. What formula would you use? thanks

    5) y
    = 2x, y = 2, x = 0
    The two "y=" is because they are trying to describe to you the region which to rotate.

    On a graph, draw the horizontal line at y=2, and shade everything below it.
    Next, draw the vertical line at x=0 and shade everything to right (positive).
    Next, draw the line y=2x and shade everything above it. This triangular region that results from the overlap of all three shaddings is the region we are rotating.
    \int_{0}^{1} \pi (2^2 - (2x)^2) \, dx

    Hopefully that gets you started. I think if you figure out the correct region you'll probably be able to get going from there. If not, let me know.
    Last edited by n0083; March 29th 2009 at 08:54 PM. Reason: upper limit of integration from 4 to 1
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  3. #3
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    Can you give me the general formula to use for any problem? and why is 2^2 at the beggining?

    question number 6 is ) y = x^2 + 5, y = 2x + 5

    would it be the integral of (x^2+5)^2 - (2x+5)^2?


    Last edited by mr fantastic; March 30th 2009 at 02:48 AM. Reason: Merged posts
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  4. #4
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    Quote Originally Posted by Freemymind View Post
    Can you give me the general formula to use for any problem? and why is 2^2 at the beggining?
    In general when rotating f(x) about the x-axis we have

    \int_a^b \pi f^2(x) \, dx

    So we are first rotating f_1(x) = 2 about the x axis which gives

    \int_a^b \pi 2^2 \, dx (*)

    We then are subtracting this volume from f_2(x) = 2x rotated about the x-axis
    \int_a^b \pi (2x)^2 \, dx (**)

    Subtracting (**) from (*) gives the previous result.
    Last edited by n0083; March 29th 2009 at 09:12 PM. Reason: meant to do preview not submit
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  5. #5
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    Quote Originally Posted by Freemymind View Post
    question number 6 is ) y = x^2 + 5, y = 2x + 5

    would it be the integral of (x^2+5)^2 - (2x+5)^2?
    If you graph these functions you will see y_1=2x+5 > y_2=x^2+5 on the x-interval [0,2]. (set the two y values to be the same and you will find the limits of integration 0 and 2)

    So it would be reversed

    \int_{0}^{2} \pi ((y_1)^2 - (y_2)^2) \, dx
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  6. #6
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    Ok so in general when you have two "Y =" functions you will find the greater one then you minus the 2nd one?
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