# Thread: Volume of Solid of revolution

1. ## Volume of Solid of revolution

Hi guys,

I'm wondering if someone can help me out with a couple of questions.

1) Find the total area enclosed by $f(x)=4x-x^2$ and the x axis.

Now, I know this involves taking an integral, but I'm unsure of what limits I'm supposed to use since it's talks about the x axis in it's entirety.

Integrating the above gives $2x^2-x^3/3+constant$ but I'm just not sure where to go with that.

2) Find the total mass (M) of a spherical cap of uniform density (density = 1) of height b cut from a sphere of radius a, where a>b.

Since the density is 1, this means it's just a volume of the solid of revolution, correct? So that means you'd take the formula for the volume of a sphere, square it and integrate it from a to b and then multiply by pi, since that's how my textbook says to do the volume.

However I'm having trouble coming up with the function...

Any help is greatly appreciated.
Thanks!

2. Hint for #1: find the roots of $f(x)$. Plot the function. See if you see where to go next.

For #2, you sound a bit confused there. You are correct in thinking that because the density is uniform, the problem reduces down to finding the volume, since the density is 1. (What are its units?) However, your method of finding the volume seems a bit off to me.

First off, whenever I see circles or spheres, I try, whenever possible, to put the origin at the center. That just makes computations easier, usually. As your thread title indicates, this is a volume of the solid of revolution. Let's say that the sphere is chopped off at some point on the x axis. In other words, the sphere is "on its side", so to speak. This makes the integral not have to be pieced together in the variable of integration. Now, what shape, when revolved around the x axis, gives you a sphere? Can you find the equation of that shape? Now use the Calc II method: chop up your problem into little bits, solve the problem there, and integrate to find the answer.

3. Originally Posted by Ackbeet
Hint for #1: find the roots of $f(x)$. Plot the function. See if you see where to go next.
Thanks for your reply. That's so obvious now. I'm so worn out from studying it seems I'm just missing the obvious now.

For #2, you sound a bit confused there. You are correct in thinking that because the density is uniform, the problem reduces down to finding the volume, since the density is 1. (What are its units?) However, your method of finding the volume seems a bit off to me.
No units mentioned. I copy+pasted the question exactly as it appears on the sheet.

First off, whenever I see circles or spheres, I try, whenever possible, to put the origin at the center. That just makes computations easier, usually. As your thread title indicates, this is a volume of the solid of revolution. Let's say that the sphere is chopped off at some point on the x axis. In other words, the sphere is "on its side", so to speak. This makes the integral not have to be pieced together in the variable of integration. Now, what shape, when revolved around the x axis, gives you a sphere? Can you find the equation of that shape? Now use the Calc II method: chop up your problem into little bits, solve the problem there, and integrate to find the answer.
A circle revolved around the axis gives a sphere, right?

4. A circle is it. What's the equation of a circle at the origin of radius a?

5. Originally Posted by Ackbeet
A circle is it. What's the equation of a circle at the origin of radius a?
Umm, well the general form of a circle is $x^2+y^2=r^2$. In this case, r=a, so we'd have $x^2+y^2=a^2$?

6. Correct. Now, given the way we've oriented the circle on the x axis, for which variable do you need to solve that equation? (Think about how you're planning on dicing up the lopped-off sphere).

7. Originally Posted by Ackbeet
Correct. Now, given the way we've oriented the circle on the x axis, for which variable do you need to solve that equation? (Think about how you're planning on dicing up the lopped-off sphere).
I'm pretty sure we'd have to solve for $y$?

8. Why is that? Justify your thinking here.

9. Originally Posted by Ackbeet
Why is that? Justify your thinking here.
Because in this case, y is a function of x since x and a will determine how it high up the cap extends. Not the other way around. I'm fairly sure. :S

10. Ok. I would plan on using the disk method for this problem, which means that the radius I need for each disk is simply y. Therefore, I have to solve for y. What do you get for that?

11. Originally Posted by Ackbeet
Ok. I would plan on using the disk method for this problem, which means that the radius I need for each disk is simply y. Therefore, I have to solve for y. What do you get for that?
We only need to solve for $y^2$ since we're going to be squaring it anyway, so we have $y^2=a^2-x^2$. $y$ would just be the square root of that.

12. Great. You're ahead of me. So now, what is the volume of one of the little disks?

13. Originally Posted by Ackbeet
Great. You're ahead of me. So now, what is the volume of one of the little disks?
Umm....

Well, we'll have the one of the discs will be the integral of $y^2$ from above. Which I believe is $integral (a^2-x^2) dx = a^2 x-x^3/3+constant$

14. Don't forget the $\pi$. I think also you need to include some limits on the integral; this is how you will take into account that the sphere has been lopped off.

15. Originally Posted by Ackbeet
Don't forget the $\pi$. I think also you need to include some limits on the integral; this is how you will take into account that the sphere has been lopped off.
Ah of course. We'd integrate from the origin to b, yes?

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