I am trying to find the volume of a cone using integration through horizontal slicing. The cone has a base radius of 10cm and a height of 5cm.

I am assuming this means I should integrate with respect to y, but I am not entirely sure how to set this up. I know that volume of a cylinder is given by the following: $$V = \pi r^2h$$

So I am assuming that the integral would be: $$\pi \int_0^5 f(y)^2dy$$

I am not sure how the x value of the radius 10cm (since it is not with respect to y) should fit into the equation, though.

Also, sorry for the pseudo-code style. I do not know how to use the math typesetting yet.


Place your cylinder so the center of the base is at $(0,0)$, and the apex is at $(0,5)$.

If you imagine looking at the cylinder straight on, it will look like a triangle with base $20$ and height $5$. If you make a horizontal slice at level $y$, then you get a figure with two similar triangles:

                   ^                  ^
                  / \                 | 
                 /   \                |
    ^           /_____\               5 
    |          /   2r  \              |
    y         /         \             |
    |        /           \            |
    V       /_____________\           V

            |----- 20 -----|

then using similar triangles note that the height of the triangle on the top is $5-y$, and the base is $2r$. So we have $$\frac{2r}{20} = \frac{5-y}{5}.$$ From this, we can express $r$ in terms of $y$.

This is where you are using the fact that the base of your original cone is $10$ cm.

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    $\begingroup$ Sorry, I am a bit thick-headed here. I understand how you derived 5-y, but I am not sure how you derived the above equality (specifically how you arrived at the ratio of r/10). Can you explain a bit more? $\endgroup$ – dtg Nov 15 '11 at 5:40
  • $\begingroup$ @Dylan: Does the picture help? $\endgroup$ – Arturo Magidin Nov 15 '11 at 5:50
  • $\begingroup$ OK. That makes sense. So solving for y from the above, i get y = -1/2r + 5. This should be plugged into the function f(y)dy i am assuming and then integrated (where b-a -> 5-0)? $\endgroup$ – dtg Nov 15 '11 at 5:57
  • $\begingroup$ @Dylan: You don't want to solve for $y$, you want to solve for $r$: the volume of the cylindrical slice at level $y$ is $\pi r^2\Delta y$, so you want to express $r$ in terms of $y$, not the other way around. $\endgroup$ – Arturo Magidin Nov 15 '11 at 6:02
  • $\begingroup$ Hmm, but when i solve the above equality for r then i get 10 - 2y. I am not sure this agrees with Andre's response above (or at least I don't see how it does - it appears he is taking the slope delta y over delta x??) $\endgroup$ – dtg Nov 15 '11 at 6:17

You can set things up so that you integrate with respect to $x$ or you can set things up so that you integrate with respect to $y$. It's your pick! For each solution, you should draw the picture that goes with that solution.

With respect to $x$: Look at the line that passes through the origin and the point $(5,10)$. Rotate the region below this line, above the $x$-axis, from $x=0$ to $x=5$, about the $x$-axis. This will generate a cone with base radius $10$ and height $5$. The main axis of this cone is along the $x$-axis. Kind of a sleeping cone.

Take a slice of width "$dx$" at $x$, perpendicular to the $x$-axis. The ordinary name for this would be a vertical slice.

This slice is almost a very thin cylinder: if you are hungry, think of a thin ham slice taken from a conical ham. Let us find the radius of this slice. The line through the origin that goes through $(5,10)$ has slope $2$, so has equation $y=2x$. Thus at $x$ the radius of our almost cylinder is $2x$. It follows that the thin slice the slice has (almost) volume $\pi(2x)^2 dx$. "Add up" (integrate) from $x=0$ to $x=5$. The volume of our cone is equal to $$\int_{x=0}^5 \pi(2x)^2 \,dx=\int_0^5 4\pi x^2\,dx.$$ The integration is easy. We get $\dfrac{500\pi}{3}$.

With respect to $y$: It is a matter of taste whether our cone is point up or point down. Since an answer with point up has already been posted, we imagine the cone with point down at the origin. Look at the line that goes through the origin and passes through the point $(10,5)$. Take the region to the left of this line, to the right of the $y$-axis, from $y=0$ to $y=5$. Rotate this region about the $y$-axis. We get a cone with base radius $10$ and height $5$.

Take a horizontal slice of width "$dy$" at height $y$. This looks almost like a flat cylindrical coin.

We want to find the volume of that coin. The line through the origin and $(10,5)$ has slope $1/2$, so it has equation $y=x/2$. So $x=2y$, and therefore the radius of our thin slice is $\pi(2y)^2 dy$. Thus the volume of the cone is $$\int_{y=0}^5 \pi(2y)^2\,dy.$$ This is the same definite integral as our previous one. Only the name of the variable of integration has changed. Naturally, the result is the same.

  • $\begingroup$ Andre, which is more easy/reasonable deriving the general formula and then substituting the values or deriving for the particular values? ... also why use cylinders when you can simply use discs? $\endgroup$ – Quixotic Nov 15 '11 at 6:09
  • $\begingroup$ Indeed, the general formula is just as easy to derive. However, sometimes concrete specific numbers can help the initial understanding. After that, the work with $r$ and $h$ seems reasonable. As to why not circle, the cross-sections are indeed circles. But I wanted to convey, unfortunately without a picture, that we are adding up the volumes of very thin disks. If one practices that enough times, the intuition behind some formulas of say Physics becomes clearer. $\endgroup$ – André Nicolas Nov 15 '11 at 6:17
  • $\begingroup$ Aha..., I got it!, I guess in these kinds of solids thinking about the volume interpretation is better than the discs however they might lead to the same thing but the volume interpretation is much more neat, another example would be while computing the volume of a paraboloid. $\endgroup$ – Quixotic Nov 15 '11 at 6:34
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    $\begingroup$ Or total displacement as definite integral of velocity, or fluid pressure, or moment about the $y$-axis, or many other things. $\endgroup$ – André Nicolas Nov 15 '11 at 6:58

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