# Homework Help: Volume of a bead

1. Sep 11, 2006

### mbrmbrg

We're trying to maximize (then minimize) the volume of a sphere wuth a cylinder drilled through the center.

So I have a circle with the equation x^2 + y^2 = R^2, and I'm going to rotate it around the y-axis. Actually, I'm only going to rotate the quarter of it that lies in the first quadrant, then multiply its volume by 2.

Now since the sphere has a cylindrical hole in it, my quarter-circle that I will rotate starts at sqrt(R^2-a^2), where a is half the height of the cylindrical hole.

(sorry for the mess; I don't know how to draw a graph on the computer)

SO.

My integrand is (2)(pi)(x)(L(x))=(2)(pi)(x)sqrt(R^2-x^2)
because the equation of a circle is y=sqrt(R^2-x^2)

My upper and lower limits are R and sqrt(R^2-a^2) respectively.

I didn't yet learn how to evaluate an integral of this form.
Is there any way to find the max/min Volumes without evaluating the integal? Or is my integral wrong altogether?

2. Sep 12, 2006

### HallsofIvy

No, that's not your integrand. You don't want to find area, you want to find the volume when the circle is rotated about an axis.

I would be very surprised if you hadn't learned to integrate such square roots long ago- its a simple trig substitution. However, as I pointed out, your integral is altogether wrong. $y= \sqrt{R^2- x^2}$ is a radius of the disk formed when the circle is rotated around the x-axis. The area of such a disk is $\pi y^2$ and the volume is
$$\pi\int y^2dx$$

3. Sep 12, 2006

### mbrmbrg

Thank you.

My Calc II professor loves his subject. Hence our starting with a brief review of the fundemental theorem (the last thing we did in calc I) and moving straight to a day on Areas, then spending time on volumes.
He says we'll get to methods of integration soon.

4. Sep 12, 2006

### mbrmbrg

My professor strongly suggested using the cylindrical shell method, which is how I got my original integrand.

In my diagram, I set R to be the radius, so wouldn't $$y\pm \sqrt{R^2-x^2}$$ be the equation of my circle (rather than a radius thereof)?

Basically, I tried doing it with the disk method and ended up with $$\pi\int \sqrt{R^2-x^2}^2dx$$ which simplifies to $$\pi\int \sqrt{R^2-x^2}dx$$, which I still can't evaluate because my class skips around. But is this the integrand I should be looking for?

Last edited: Sep 12, 2006
5. Sep 12, 2006

### HallsofIvy

No, it doesn't! Squaring a squareroot doesn't leave a squareroot! That simplifies to
$$\pi \int (R^2- x^2) dx$$

6. Sep 12, 2006

### mbrmbrg

Would you believe me if I told you that was a typo? Because it was

I'm sorry to be so dense, but I looked it over yet again and still can't follow your reasoning. Doesn't $$A(x)=\pi R^2$$ rather than $$A(x)=\pi y^2$$?
Why are we putting y (in the form $$\sqrt{R^2-x^2}$$) into $$\int (A(x)) dx$$ instead of putting in R (in the form $$\sqrt{y^2+x^2}$$)?
Other than the fact that an integral with x's and y's looks really ugly, that is...

7. Sep 12, 2006

### mbrmbrg

yowzers, naps should be mandatoory!
You're rotating around the xaxis, aren't you? that would make y the radius of that disk... Good Lord. Sheesh.
Thanks, and good morning!