1. [SOLVED] Tricky volume problem

What volume is left of a cone with the radius $\displaystyle 2a$ and the height $\displaystyle h$, when you cut out a cylinder with the radius $\displaystyle a$ 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:
$\displaystyle \sqrt{x^2 + y^2} \leq z \leq h$

And the cylinder's projection on the xy-plane as:
$\displaystyle D= \left\{(x,y)\in R^2: (x-a)^2 + y^2 \leq a^2 \right\}$

The volume of the untouched half of the cone is already known: $\displaystyle \frac{2}{3}\pi r^2 h$

Integrating all this is not easy though:
$\displaystyle \int \int_D (h - \sqrt{x^2 + y^2})dxdy$

By making the variable change $\displaystyle u = x - a,\ v = y$ 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:
$\displaystyle \int \int_E (h - \sqrt{(u+a)^2 + v^2})dudv$

Especially in polar coordinates:
$\displaystyle \int \int_F (h - \sqrt{(rcos\varphi+a)^2 + r^2sin^2\varphi})r drd\varphi$

Is there an easier way to solve this problem?

2. simplify your function by using some factoring
remember cos^2(x) + sin^2(x) = 1

3. There is supposely an easier way to solve this problem, but I can't think of anything. Anyone?

4. Originally Posted by Spec
What volume is left of a cone with the radius $\displaystyle 2a$ and the height $\displaystyle h$, when you cut out a cylinder with the radius $\displaystyle a$ from the cone (so that the surface of the cylinder is touching the axis of symmetry of the cone.
the equation of (the upper half of course) of the cone is $\displaystyle z=\frac{h}{2a}\sqrt{x^2+y^2},$ and the equation of the cylinder is $\displaystyle x^2+y^2=2ax,$ which in polar coordinates becomes:

$\displaystyle r=2a \cos \theta, \ \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}.$ thus: the volume of the cone cut by the cylinder = $\displaystyle \int \int_D (h - \frac{h}{2a} \sqrt{x^2+y^2}) \ dA = h \times (\text{the area of D}) - \frac{h}{2a}\int \int_D \sqrt{x^2+y^2} \ dA=$

$\displaystyle \pi a^2 h - \frac{h}{2a}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_0^{2a\cos \theta}r^2 \ dr \ d \theta = \pi a^2 h - \frac{4}{3}a^2 h\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos^3 \theta \ d \theta=\pi a^2 h - \frac{16}{9}a^2h.$ hence the volume left = $\displaystyle \frac{4}{3}\pi a^2 h - \left(\pi a^2 h - \frac{16}{9}a^2h \right)=\left(\frac{\pi}{3} + \frac{16}{9} \right)a^2 h. \ \ \ \square$

5. 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 $\displaystyle 2a$ and the height $\displaystyle h$,
when you cut out a cylinder with the radius $\displaystyle a$ 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 $\displaystyle 2a$ and the radius of the cylinder is $\displaystyle a,$
. . then the situation must look like this:
Code:
    -         *
:        /:\
:       / : \
:      /  :  \
h:     *---+---*
:    /|   :   |\
:   / |   :  y| \
:  /  |   :   |  \
- *---*---+---*---*
a   a   a   a

Let the height of the cylinder by $\displaystyle y.$

From similar right triangles, we have: .$\displaystyle \frac{y}{a} \:=\:\frac{h}{2a} \quad\Rightarrow\quad y \:=\:\frac{h}{2}$

The volume of a cone is: .$\displaystyle V \:=\:\frac{1}{3}\pi r^2h$
. . So we have: .$\displaystyle V_{cone} \;=\;\frac{1}{3}\pi(2a)^2h \;=\;\frac{4}{3}\pi a^2h$

The volume of a cylinder is: .$\displaystyle V \:=\:\pi r^2h$

. . So we have: .$\displaystyle V_{cyl} \:=\: \pi a^2y \:=\:\pi a^2\left(\frac{h}{2}\right) \:=\:\frac{1}{2}\pi a^2h$

Therefore, the remaining volume is: .$\displaystyle \frac{4}{3}\pi a^2h - \frac{1}{2}\pi a^2h \;=\;\boxed{\frac{5}{6}\pi a^2h}$

6. 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 $\displaystyle z=\frac{h}{2a}\sqrt{x^2+y^2},$ and $\displaystyle x^2+y^2=2ax,$ come from.

7. Originally Posted by Soroban

If the radius of the cone is $\displaystyle 2a$ and the radius of the cylinder is $\displaystyle a,$
. . then the situation must look like this:
Code:
    -         *
:        /:\
:       / : \
:      /  :  \
h:     *---+---*
:    /|   :   |\
:   / |   :  y| \
:  /  |   :   |  \
- *---*---+---*---*
a   a   a   a

Soroban, I think it should be something like this:
Code:

-         *
:        /:\
:       / : \
:      /  :  \
h:     *   +---*---+
:    /    |    \  |
:   /     |     \ | y
:  /      |      \|
- *---*---+---*---*
a   a   a   a

@Spec:

$\displaystyle z=h-\frac{h}{r}\sqrt{x^2+y^2}$ is the equation of a cone of radius r and height h.

$\displaystyle x^2+y^2 = 2ax$ can be derived from $\displaystyle (x-a)^2+y^2 =a^2$, which is the equation of the cylinder.

8. Hi guys. Thought I'd try and illustrate what I think NonCommAlg is doing above. The integral $\displaystyle \iint_D \left(h-\frac{h}{2a}\sqrt{x^2+y^2}\right)dA$ 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 $\displaystyle dA$ over the region $\displaystyle D$ with $\displaystyle D$ 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 $\displaystyle 4/3\pi a^2 h$
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}]

9. Originally Posted by wingless
$\displaystyle z=h-\frac{h}{r}\sqrt{x^2+y^2}$ is the equation of a cone of radius r and height h.
Where is that equation derived from?

10. Originally Posted by Spec
Where is that equation derived from?
It can be derived easily.

First think of the equation $\displaystyle z=\sqrt{x^2+y^2}$. 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, $\displaystyle -\frac{h}{r}\sqrt{x^2+y^2}$. And add h to it in order to place the vertex at (0,0,h). There, you have it: $\displaystyle h-\frac{h}{r}\sqrt{x^2+y^2}$

11. Originally Posted by NonCommAlg
$\displaystyle x^2+y^2=2ax,$ which in polar coordinates becomes:

$\displaystyle r=2a \cos \theta, \ \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}.$
I don't understand that part.

$\displaystyle x = r cos\theta$ in polar coordinates, and the radius is $\displaystyle \sqrt{2ax}$ in cartesian coordinates and should thus be $\displaystyle \sqrt{2a \cdot rcos\theta}$ in polar coordinates.

And why do you only integrate one half of the circle?

12. Originally Posted by Spec
I don't understand that part.

$\displaystyle x = r cos\theta$ in polar coordinates, and the radius is $\displaystyle \sqrt{2ax}$ in cartesian coordinates and should thus be $\displaystyle \sqrt{2a \cdot rcos\theta}$ in polar coordinates.

And why do you only integrate one half of the circle?
$\displaystyle x^2+y^2=2ax ~\iff~ x^2-2ax+{\color{red}a^2} + y^2 = \color{red}a^2 ~\iff~ \boxed{(x-a)^2+y^2=a^2}$

This is the equation of the cross-section of the cylinder. It is a circle with the center C(a, 0) and the radius r = a. For each pair (x, y) you have $\displaystyle z \in \mathbb{R}$

13. Originally Posted by earboth
$\displaystyle x^2+y^2=2ax ~\iff~ x^2-2ax+{\color{red}a^2} + y^2 = \color{red}a^2 ~\iff~ \boxed{(x-a)^2+y^2=a^2}$

This is the equation of the cross-section of the cylinder. It is a circle with the center C(a, 0) and the radius r = a. For each pair (x, y) you have $\displaystyle z \in \mathbb{R}$
I understand that part, but I don't understand how he got that radius in polar coordinates. My calculations show that it should be $\displaystyle \sqrt{2ax}$ in carthesian coordinates and $\displaystyle \sqrt{2a \cdot r cos\theta}$ in polar coordinates.

14. Originally Posted by Spec
I understand that part, but I don't understand how he got that radius in polar coordinates. My calculations show that it should be $\displaystyle \sqrt{2ax}$ in carthesian coordinates and $\displaystyle \sqrt{2a \cdot r cos\theta}$ in polar coordinates.
1. to help you see why the equation of the cone is $\displaystyle z=\frac{h}{2a}\sqrt{x^2+y^2},$ for example look at the image of the cone in the first quadrant of $\displaystyle yz$ plane, which is the line $\displaystyle z=\frac{h}{2a}y.$

2. in polar coordinates we have $\displaystyle x=r\cos \theta$ and $\displaystyle x^2+y^2=r^2.$ thus $\displaystyle x^2+y^2=2ax$ becomes $\displaystyle r^2=2ar\cos \theta,$ which, after dividing both sides by $\displaystyle r,$ gives us $\displaystyle r=2a\cos \theta.$

the easiest way to see why $\displaystyle \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}$ is to draw the circle!