simplify your function by using some factoring
remember cos^2(x) + sin^2(x) = 1
What volume is left of a cone with the radius and the height , when you cut out a cylinder with the radius from the cone (so that the surface of the cylinder is touching the axis of symmetry of the cone.
The volume of the cone can be described as:
And the cylinder's projection on the xy-plane as:
The volume of the untouched half of the cone is already known:
Integrating all this is not easy though:
By making the variable change we get a projection on the xy-plane that can easily be turned into polar coordinates, but integrating this turned out to be quite difficult:
Especially in polar coordinates:
Is there an easier way to solve this problem?
Hello, Spec!
Please check the wording of the problem.
. . As stated, the situation is impossible.
Also, we don't need Calculus. . .
What volume is left of a cone with the radius and the height ,
when you cut out a cylinder with the radius from the cone so that
the surface of the cylinder is touching the axis of symmetry of the cone. .??
If the radius of the cone is and the radius of the cylinder is
. . then the situation must look like this:Code:- * : /:\ : / : \ : / : \ h: *---+---* : /| : |\ : / | : y| \ : / | : | \ - *---*---+---*---* a a a a
Let the height of the cylinder by
From similar right triangles, we have: .
The volume of a cone is: .
. . So we have: .
The volume of a cylinder is: .
. . So we have: .
Therefore, the remaining volume is: .
The radius of the cone at the base is 2a. NonCommAlg's answer is correct, but I have no idea how he got there.
If you split the base of the cone in half, and then drill a hole right through with the radius a into that half, you'll end up with the volume I was thinking of.
I really don't get where and come from.
Soroban, I think it should be something like this:
Code:- * : /:\ : / : \ : / : \ h: * +---*---+ : / | \ | : / | \ | y : / | \| - *---*---+---*---* a a a a
@Spec:
is the equation of a cone of radius r and height h.
can be derived from , which is the equation of the cylinder.
Hi guys. Thought I'd try and illustrate what I think NonCommAlg is doing above. The integral is that part of the blue cylinder inside of the the red cone: it's the height of the cone h, minus the intersecting surface of the cylinder with the cone times over the region with being the area of the base of the cylinder. Guess that's confusing though. But once you see that, then it makes sense once you figure what that volume of blue inside the cone is, then just subtract that volume from the total volume of the cone which is
Here's the Mathematica code for the first figure in case some are interested (the code for the other two is a bit messy):
Code:a = 1/2; h = 1; cone = RevolutionPlot3D[t/a, {t, 0, a}, BoxRatios -> {1, 1, 1}, PlotStyle -> Red] cy = Graphics3D[{Blue, Cylinder[ {{0, a/2, 0}, {0, a/2, h}}, a/2]}] Show[{cone, cy}]
It can be derived easily.
First think of the equation . Draw it on the xz plane. Do you see the isosceles right triangle? Stretch it by h/r. Now it has the right proportion. Take the negative of it, . And add h to it in order to place the vertex at (0,0,h). There, you have it:
1. to help you see why the equation of the cone is for example look at the image of the cone in the first quadrant of plane, which is the line
2. in polar coordinates we have and thus becomes which, after dividing both sides by gives us
the easiest way to see why is to draw the circle!