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Thread: 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

    $\displaystyle x^2+y^2 = R^2$

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

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

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

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

    The volume is then

    $\displaystyle V = 2L\int_{-R}^{h-R} \sqrt{R^2-y^2}\; dy$.
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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. $\displaystyle h $ is a measure from the bottom of the cylinder to some height $\displaystyle h$. If $\displaystyle h = 0$ then

    $\displaystyle
    V = 2L\int_{-R}^{-R} \sqrt{R^2-y^2}\; dy = 0
    $

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

    $\displaystyle
    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}
    $
    which is the area of half a circle times L.
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