Maximizing/Minimizing Volume of a Sphere with Cylinder Drilled Through Center

[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?

 

[HallsofIvy]

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)

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.

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?

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
$$\pi\int y^2dx$$

 

[mbrmbrg]

Thank you.

My Calc II professor loves his subject. Hence our starting with a brief review of the fundamental 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.

 

[mbrmbrg]

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.

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

$y= \sqrt{R^2- x^2}$ 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
$$\pi\int y^2dx$$

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?

 

[HallsofIvy]

Basically, I tried doing it with the disk method and ended up with $$\pi\int \sqrt{R^2-x^2}^2dx$$ which simplifies to $$\pi\int \sqrt{R^2-x^2}dx$$

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

 

[mbrmbrg]

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

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...

 

[mbrmbrg]

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...

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!