A hole of radius r is bored through a cylinder of radius R at right angles

to the axis of the cylinder. Set up, but do not evaluate, an integral for the

volume cut out.

Printable View

- May 28th 2008, 12:45 PMevilpostingmongyVolume help!
A hole of radius r is bored through a cylinder of radius R at right angles

to the axis of the cylinder. Set up, but do not evaluate, an integral for the

volume cut out. - May 28th 2008, 01:22 PMgalactus
Looking down the 'throat' of the cylinder along the axis, we have something that looks kind of like this.

From the diagram, $\displaystyle y=\sqrt{R^{2}-x^{2}}$

But x=r. So, $\displaystyle y^{2}=R^{2}-r^{2}$

$\displaystyle x=\sqrt{R^{2}-y^{2}}$

$\displaystyle {\pi}\int_{-\sqrt{R^{2}-r^{2}}}^{\sqrt{R^{2}-r^{2}}}\left[(R^{2}-y^{2})-r^{2}\right]dy$

How's that?. - May 28th 2008, 01:24 PMgalactus
Oops, you know what I did?. I gave the volume of the solid after you drill the hole, not the hole itself(Headbang). Can you adjust accordingly looking at the diagram or just subtracting it from the volume of the entire cylinder?.

- May 28th 2008, 01:41 PMevilpostingmongy
The thing that gets me is that the books answer makes no sense to me.

8integralsign(lim r to 0) sqrt(R^2-y^2)*sqrt(r^2-y^2)dy

that 8 is not a typo I have no I idea how that constant got there.[ - May 28th 2008, 02:24 PMevilpostingmongy
The way I'd do it, I'd take a parabola y=R-x^2 and rotate about the y axis with the limits of integration x= r to x=0. so S (S is the integral sign) r to 0

pi(R-x^2)^2dx but that is way different from the answer I posted. - May 28th 2008, 02:29 PMevilpostingmongy
I have a plan. I see a rectangle in that drawing. I will tackle this problem when I'm done with dinner. Don't post any solutions until I am done with the

problem (hate to be a board fascist lol). - May 28th 2008, 02:35 PMgalactus
I see what they done.

$\displaystyle 8\int_{0}^{r}\sqrt{R^{2}-y^{2}}\sqrt{r^{2}-y^{2}}dy$

They broke the drilled hole up into octants(hence the multiplication by 8), then multiplied the area of the base by the height.

The height being

$\displaystyle \sqrt{R^{2}-y^{2}}$

and the area of the base being

$\displaystyle \sqrt{r^{2}-y^{2}}$.

Of course, then they integrated over the region that makes up the hole.

That particular integral is a booger. I can see why they said do not evaluate. - May 28th 2008, 02:48 PMgalactus
Actually, thinking about it, what I worked out is for a hole through a sphere instead of a cylinder. It gives the volume remaining after a hole is drilled through a sphere. You can keep it if you wish or I can delete it.

- May 28th 2008, 03:27 PMevilpostingmongy
Okay I took the x section to be a rectangle. The width of the rectangle

is a piece of a circle with respect to y so it is sqrt(R^2-y^2) with R being

the radius of the cylinder. The length of the rectangle is the y of the

piece of the hole (which is also circular) with the equation of sqrt (r^2-y^2).

The area of the rectangle varies along the radius of the holes so

the two circular functions are put in l*w form. What confuses me is why the

8 is there. - May 28th 2008, 03:30 PMevilpostingmongy
Oh I typed that up without reading your explaination. Oh and octants?

Why did they do that? - May 28th 2008, 03:45 PMgalactus
Because it is easier a lot of times. You see it a lot when you use double or triple integration. That is a Calc III topic though.