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Math Help - proof of the equation of the volume of a cylinder

  1. #1
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    Lightbulb proof of the equation of the volume of a cylinder

    hey everyone!
    can someone please prove the formula for the volume of a cylinder for me?
    thank you so much!
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    Re: proof of the equation of the volume of a cylinder

    A prism is a 3D shape will all identical cross-sections.

    A cylinder is a prism with circular cross-sections.

    The volume of a prism is the cross-sectional area X height of the prism (in other words, the number of cubes in one layer - which is numerically the same as the cross-sectional area, times the number of layers - which is numerically the same as the height of the prism).

    Go from here.
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    Re: proof of the equation of the volume of a cylinder

    ^what Prove It said
    For prisms, V=Bh.
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    Re: proof of the equation of the volume of a cylinder

    We have:

    $\displaystyle V = \int_{-r}^r \int_{-\sqrt{r^2 - x^2}}^{\sqrt{r^2 - x^2}} \int_0^h \ dz\ dy\ dx$

    $\displaystyle = \int_{-r}^r \int_{-\sqrt{r^2 - x^2}}^{\sqrt{r^2 - x^2}} h\ dy\ dx$

    $\displaystyle = h \int_{-r}^r \int_{-\sqrt{r^2 - x^2}}^{\sqrt{r^2 - x^2}} \ dy\ dx$

    $\displaystyle = h \int_{-r}^r 2\sqrt{r^2 - x^2}\ dx$

    $\displaystyle = 2h \int_{-r}^r \sqrt{r^2 - x^2}\ dx$

    Let $x = r\sin u$ so that $dx = r\cos u\ du$. As $x$ goes from $-r$ to $r$, $u$ goes from $-\dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$.

    $\displaystyle = 2h \int_{-\pi/2}^{\pi/2} \sqrt{r^2 - r^2\sin^2 u}\ r\cos u\ du$

    $\displaystyle = 2r^2h \int_{-\pi/2}^{\pi/2} \cos^2 u\ du$

    $\displaystyle = 2r^2h \int_{-\pi/2}^{\pi/2} \dfrac{1 + \cos 2u}{2}\ du$

    $\displaystyle = r^2h \int_{-\pi/2}^{\pi/2} 1 + \cos 2u\ du$

    $\displaystyle = r^2h \left(\int_{-\pi/2}^{\pi/2}\ du + \int_{-\pi/2}^{\pi/2} \cos 2u\ du\right)$

    $\displaystyle = r^2h \left(\dfrac{\pi}{2} - \left(-\dfrac{\pi}{2}\right) + \dfrac{1}{2}\int_{-\pi/2}^{\pi/2} \cos 2u\ 2du\right)$

    $= \pi r^2h + \dfrac{r^2h}{2}(\sin(\pi) - \sin(-\pi))$

    $= \pi r^2h + \dfrac{r^2h}{2}(0)$

    $= \pi r^2h$.
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    Re: proof of the equation of the volume of a cylinder

    Quote Originally Posted by Deveno View Post
    We have:

    $\displaystyle V = \int_{-r}^r \int_{-\sqrt{r^2 - x^2}}^{\sqrt{r^2 - x^2}} \int_0^h \ dz\ dy\ dx$

    $\displaystyle = \int_{-r}^r \int_{-\sqrt{r^2 - x^2}}^{\sqrt{r^2 - x^2}} h\ dy\ dx$

    $\displaystyle = h \int_{-r}^r \int_{-\sqrt{r^2 - x^2}}^{\sqrt{r^2 - x^2}} \ dy\ dx$

    $\displaystyle = h \int_{-r}^r 2\sqrt{r^2 - x^2}\ dx$

    $\displaystyle = 2h \int_{-r}^r \sqrt{r^2 - x^2}\ dx$

    Let $x = r\sin u$ so that $dx = r\cos u\ du$. As $x$ goes from $-r$ to $r$, $u$ goes from $-\dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$.

    $\displaystyle = 2h \int_{-\pi/2}^{\pi/2} \sqrt{r^2 - r^2\sin^2 u}\ r\cos u\ du$

    $\displaystyle = 2r^2h \int_{-\pi/2}^{\pi/2} \cos^2 u\ du$

    $\displaystyle = 2r^2h \int_{-\pi/2}^{\pi/2} \dfrac{1 + \cos 2u}{2}\ du$

    $\displaystyle = r^2h \int_{-\pi/2}^{\pi/2} 1 + \cos 2u\ du$

    $\displaystyle = r^2h \left(\int_{-\pi/2}^{\pi/2}\ du + \int_{-\pi/2}^{\pi/2} \cos 2u\ du\right)$

    $\displaystyle = r^2h \left(\dfrac{\pi}{2} - \left(-\dfrac{\pi}{2}\right) + \dfrac{1}{2}\int_{-\pi/2}^{\pi/2} \cos 2u\ 2du\right)$

    $= \pi r^2h + \dfrac{r^2h}{2}(\sin(\pi) - \sin(-\pi))$

    $= \pi r^2h + \dfrac{r^2h}{2}(0)$

    $= \pi r^2h$.
    Posted in "Pre-Calculus"...
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    Re: proof of the equation of the volume of a cylinder

    Quote Originally Posted by Prove It View Post
    Posted in "Pre-Calculus"...
    *Shrug*, he asked for a proof, not a statement of fact without justification. Not my problem the idea of volume of a region needs calculus to be defined. Do you know a proof of the volume of the cylinder, that doesn't involve limits in some way? Let's hear it.
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    Re: proof of the equation of the volume of a cylinder

    Quote Originally Posted by Deveno View Post
    *Shrug*, he asked for a proof, not a statement of fact without justification. Not my problem the idea of volume of a region needs calculus to be defined. Do you know a proof of the volume of the cylinder, that doesn't involve limits in some way? Let's hear it.
    Since when does volume need calculus to be defined. The volume of a shape is the amount of space inside it (measured in cubes).

    A prism is a shape with all identical cross-sections.
    A cylinder is a prism with circular cross-sections.
    The volume of any prism is equal to the number of cubes in one layer times the number of layers.
    The number of cubes is numerically the same as the cross-sectional area. The number of layers is numerically the same as the height of the prism.
    The area of the cross section (circle) is $\displaystyle \begin{align*} \pi\,r^2 \end{align*}$.
    Thus the volume of the cylinder is $\displaystyle \begin{align*} V = A\,H = \pi\,r^2\,h \end{align*}$.
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  8. #8
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    Re: proof of the equation of the volume of a cylinder

    Last time I looked, cubes didn't fit in round holes. There's some space left-over. Of course, you could make the cubes smaller, and use more of them. But there would still be small spaces left-over, it wouldn't be exact. Oh hey, I know-we could make the cubes "infinitely small" and keep track of the ratio of the number of cubes needed to the size of the cubes. Hmm, that's a limit.

    The whole "cube analogy"/number of cubes, tends to break down with irrational numbers anyway-what does $\sqrt{2}$ cubes mean? Real numbers are limits of cauchy rational sequences. I'm sorry if this offends you.

    "Measuring space" isn't as simple as it sounds. There are two related problems: what is "space", and what is "measurement"? One has to define these terms at some point, and one has to define what it means to apply these terms to a region.

    It's not that I have any doubt of that the area of a circle of radius $r$ is $\pi r^2$. The problem is, people are told that this is true, without any justification. There is a reason for this: the justification is *complicated*. Calculating what $\pi$ is, in terms of some other numbers "we know better" (like 22 and 7, or 314,159 and 100,000) requires sophisitcated (relatively) methods.

    Likewise, measuring the area bounded by a curve, is not as simple as it sounds. What's a "curve"? What is it's boundary? How do we calculate it's area, if it even exists?

    Does the original poster have any idea of the complexity of this question? I have no idea. They may not, they may still be at the stage in their education where "proof by expert opinion" is good enough. While that be a sound social strategy, it certainly isn't mathematics. Read your own signature.
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