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Math Help - Volume problem

  1. #1
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    Volume problem

    Determine the volume generated when the ellipse \frac {x^2} {9} +y^2=1 is rotated around the line y=5

    In my solution book, it says

    Translating the ellipse -5 units along the y-axis

    Why -5? Aren't you supposed to translate it 5 units?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Translations in General

    Well You are translating up 5 units by using (y-5)

    Remeber (x-a) translates a graph a units to the right and (x+5) translates to the left 5 units.
    similarly (y-a) translates up a units and (y+a) translates down a units.

    Think of the parabola y = x^2 with vertex at (0,0)

    For y = (x-2)^2 the vertex is at (2,0) where y = 0 . A translation to the right of 2 units.

    Hope this helps.
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  3. #3
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    Quote Originally Posted by chengbin View Post
    Determine the volume generated when the ellipse \frac {x^2} {9} +y^2=1 is rotated around the line y=5

    In my solution book, it says

    Translating the ellipse -5 units along the y-axis

    Why -5? Aren't you supposed to translate it 5 units?
    Isn't there something after "translating the ellipse -5 units along the y-axis"?! Why translate at all? You don't have to translate it at all. But translating it down 5 shifts the line y= 5 down to y= 0. Instead of translating around y= 5, you will be rotating around the x-axis, which, presumably you have experience in doing that. It is not really necessary to translate at all.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Volume problem

    I think I misunderstood you original question.

    by translating down 5 units and rotating about the x-axis you don't seem to simplify the problem.

    You have x^2/9 +(y+5)^2 = 1

    solving for y the upper half is y = -5+ (1-x^2/9)^1/2 and the lower half is
    y = -5- (1-x^2/9)^1/2

    You have a washer with outer radius y = -5- (1-x^2/9)^1/2 and inner radius y = -5+ (1-x^2/9)^1/2

    Without translating you have a washer with outer radius y = 5+ (1-x^2/9)^1/2 and inner radius 5- (1-x^2/9)^1/2.

    What I'd reccomend is the theorem of Pappus V =2(pi)bA
    Where A is the Area of the ellipse pi(3)(1) and b is the distance from the
    center of mass to the line y = 5. since the center of mass is (0,0)
    You have 30 (pi)^2
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