# Find the volume of a cone using integration through horizontal slicing. The cone has a base radius of 10cm and a height of 5cm.

Finding volume of a cone through integration
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:
$V=\pi {r}^{2}h$
So I am assuming that the integral would be:
$\pi {\int }_{0}^{5}f\left(y{\right)}^{2}dy$
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.
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Step 1
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 $\pi \left(2x{\right)}^{2}dx$. "Add up" (integrate) from $x=0$ to $x=5$. The volume of our cone is equal to ${\int }_{x=0}^{5}\pi \left(2x{\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{0}^{5}4\pi {x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx.$.
The integration is easy. We get $\frac{500\pi }{3}$.
Step 2
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 $\pi \left(2y{\right)}^{2}dy$. Thus the volume of the cone is ${\int }_{y=0}^{5}\pi \left(2y{\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dy.$
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.