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Math Help - Really hard Volume Question. Can someone help please?

  1. #1
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    Really hard Volume Question. Can someone help please?

    Can someone help me with this question? I don't know where to begin.

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  2. #2
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    Quote Originally Posted by s3n4te View Post
    Can someone help me with this question? I don't know where to begin.

    We'll find the cross sectional area and multiply by L. If we assume that the circle is centered at the origin then it has the equation

    x^2+y^2 = R^2

    so the area can be found by (we'll use some symmetry)

    2\int_{-R}^{R} \sqrt{R^2-y^2}\; dy

    This gives \pi R^2 but we're not going all the way up, just to a height h from the bottom or h-R from the origin so

    2\int_{-R}^{h-R} \sqrt{R^2-y^2}\; dy,

    The volume is then

    V = 2L\int_{-R}^{h-R} \sqrt{R^2-y^2}\; dy.
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  3. #3
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    I might be wrong but I think you made a small mistake because your limits of integration don't make sense. It's negative h for the bottom half and then h for the top half. Again, I'm kinda nooby at calculus so I might be wrong...
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  4. #4
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    Quote Originally Posted by Kaitosan View Post
    I might be wrong but I think you made a small mistake because your limits of integration don't make sense. It's negative h for the bottom half and then h for the top half. Again, I'm kinda nooby at calculus so I might be wrong...
    Well, let's think about it. h is a measure from the bottom of the cylinder to some height h. If h = 0 then

    <br />
V = 2L\int_{-R}^{-R} \sqrt{R^2-y^2}\; dy = 0<br />

    If h > 0 then h-R > -R so even if h < R then h-R > -R. Consider the case where h = R (the bottom half of the circle)

    <br />
V = 2L\int_{-R}^{0} \sqrt{R^2-y^2}\; dy = \left. \frac{y \sqrt{R^2 -y^2}}{2} + \frac{R^2}{2} \sin^{-1} \frac{y}{R} \right| _{-R}^0 = 2L \cdot \frac{\pi R^2}{4} = \frac{\pi R^2 L}{2}<br />
    which is the area of half a circle times L.
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