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Math Help - Volume figure of revolution

  1. #1
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    Volume figure of revolution

    Pls, help me with this problem

    Calculate the volume of the body formed by rotating a figures (x^2)+3y=0 and y+3=0 around the Y-axis.

    Thanks!
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  2. #2
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    First thing to do is calculate your limits of integration. You can do that by finding where your curves meet.
    In the second equation, you can see that y = -3 so substitute y = -3 into the first equation and solve for x.
    x^2 + 3*-3 = 0
    x = (+/-)3, so we're integrating from x=-3 to 3.

    Since our limits of integration are in terms of x, we will have to use a method that uses an integral with respect to x.
    Rotating about the y axis & integrating with respect x means we'll be using the shell method.

    Now we know what our integral will look like: 2(pi)*INTEGRAL(r(x)h(x)dx) (from a to b)
    where r(x) is the distance from the axis of rotation and h(x) is the height at x.

    Because our axis is the y-axis, distance from it is simply 'x.'
    For h(x) we have to ask our self which curve is on top, and we observe that x^2 - 9 < -3 on the interval [-3, 3]. so our height function will be
    h(x) = (-3) - (x^2 - 9).

    Now we know all the parts needed to calculate volume:
    V = 2(pi)*INTEGRAL(x*[(-3) - (x^2 - 9)]dx) from -3 to 3.
    Last edited by m.walunga; July 21st 2009 at 05:48 AM. Reason: Its not breaking
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by m.walunga View Post

    Now we know all the parts needed to calculate volume:
    V = 2(pi)*INTEGRAL(x*[(-3) - (x^2 - 9)]dx) from -3 to 3.
    Are you sure: from -3 to 3?

    Maybe: from 0 to 3?
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  4. #4
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    I'm almost certain, maybe someone else can set the record on the matter. However, I think if the integrand is even, then you can make it 4(pi)*(integral from 0 to 3)
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Pls, help me with this problem

    Calculate the volume of the body formed by rotating a figures (x^2)+3y=0 and y+3=0 around the Y-axis.

    Thanks!
    Look this picture



    I think that it is better to shift the body of rotation for three units up along the Y-axis, i.e. in the new coordinate system {x^*} = x<br />
and {y^*} = y + 3 we have y_1^* = {y_1} + 3 =  - 3 + 3 = 0 and y_2^* = {y_2} + 3 =  - \frac{{{x^{*2}}}}{3} + 3 = 3 - \frac{{{x^{*2}}}}{3}.

    Then {V_y} = 2\pi \int\limits_0^3 {{x^*}\left( {3 - \frac{{{x^{*2}}}}{3}} \right)d{x^*}}  =  \ldots  = \frac{{27}}{2}\pi {\text{ }}\left( {{\text{cubic units}}} \right).
    Last edited by Krizalid; July 21st 2009 at 06:50 AM.
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