Volume of a Sphere
This demonstration nicely explains the reasoning behind the formula for the volume of a sphere. I just fiddle with it here to perhaps make it clearer.
The big idea behind the demonstration is this: Suppose you have two different solid figures, both with the same height and the same width sitting side by side. Choose a height and make a horizontal slice at that height through both solid figures. If you can show that the resulting cross sections have the same area no matter where you make that slice, then the two solid figures MUST have the same volume.
We start with a cone whose height and radius are the same (left figure). We'll label both the height and the radius with b. Now imagine enlarging this cone proportionally to form the larger cone you see at the right. Its height and radius are similarly equal. We'll label the height and radius of the larger cone with r.
One very important idea to “see” before we move on. The area of the circular base of the smaller cone can be represented as &pib2. And no matter how we shrink or enlarge this smaller cone (proportionally), its base will ALWAYS have an area of &pib2.
Without stopping for a moment to think about that, it may sound ridiculous. Of course the circular base is always going to have an area of &pib2. The radius is, after all, b.
But not so fast. The height is also b, so what we’re saying is that no matter what the height of the smaller cone is, its circular base will ALWAYS have a radius equal to the height of the cone. Thus, the smaller cone will always have an area of &pib2. In other words, we know that the area of the base is a function of its radius (because we're using only positive values of b). But in this case, we've set it up so that the area of the base is a function of its height.
Okay, so let's take this bi-conical construction and put it inside a cylinder with the same radius and height as the larger cone.
Now let's make a horizontal slice all the way through the larger cone and the cylinder. We'll make our slice right at the circular base of the smaller cone.
If we took this cross section out of the cylinder, we would see a perfect circle, and that circle would have a radius of r. Thus, the circle would have an area of &pir2. In fact, since the radius of the cylinder remains the same up and down its height, we would see the same circle with the same area no matter where we made our slice. Thus, no matter where we make our slice, the resulting circular cross section would have an area of &pir2.
Now take a look at the cross section above. If we take that cross section and remove the area of the circular base of the smaller cone, what would the area of the cross section be?
Well, that's easy. Since the area of the circular base of the smaller cone is &pib2, we just subtract that area from the total area of the circular cross section. The result is &pir2 &minus &pib2.
Here's the kicker. We explained that no matter where we make the slice through the cylinder, the resulting circular cross section would have an area of &pir2. And we also explained that no matter where we make the slice through the cylinder, the area of the circular base of the cone would have an area of &pib2. THEREFORE, no matter where we make our slice, the resulting cross section without the area of the circular base of the cone will have an area of &pir2 &minus &pib2. Using the Distributive Property to rewrite this, we get &pi × (r2 &minus b2).
Okay, now let's construct a hemisphere (a half-sphere) with the same radius and height as the cylinder. We'll make a slice through this hemisphere at the same height (b) as we made our slice through the cylinder to form another circular cross section.
Now take a look at the right triangle formed in the hemisphere. The hypotenuse of this triangle is equal to r, because all the points that form a hemisphere are an equal distance from a fixed center point. And we see that one of the legs of this right triangle has a length of b. We can also see that the side length that we don't know is the radius of the circular cross section.
We can't say that this side length is r, because it's slightly smaller than that. As you can see, the "higher" we make our slice through the hemisphere, the smaller the resulting circular cross section would be. Similarly, the "lower" we make our slice, the larger the resulting cross section would be. Only at the base of the hemisphere would the "cross section" have a radius of r. Otherwise, it's going to be smaller than r.
So, what is this side length, this radius of the circular cross section in the hemisphere? Well, as it turns out, we don't need to find this side length. We can just figure out the square of this side length. Using the Pythagorean equation, we see that the square of the missing side length is equal to r2 &minus b2.
So since the area of the circular cross section of the hemisphere is equal to &pi × radius2, and radius2 in this case is r2 &minus b2, we can write the area of the cross section of the hemisphere as &pi × (r2 &minus b2).
This, of course, is the exact same area as the cross section in the cylinder at the same height with the area of the circular base of the smaller cone removed. And, just as it was for the cone inside the cylinder, the area of the circular cross section in the hemisphere is a function of its height. (The Pythagorean equation will not change. No matter where the cross section occurs, the radius will be "hypotenuse squared minus length of leg squared.") Thus, no matter where we slice through the hemisphere, the resulting cross section will ALWAYS have an area of &pi × (r2 &minus b2). Here are the two figures again.
Both the cylinder and the hemisphere have the same height and the same radius. And at every point along their heights, the circular cross sections have the same area so long as we remove the area of the base of the cone in the cylinder.
What this means is this: The volume of a hemisphere is equal to the volume of a cylinder with the same height and radius minus the volume of a cone with the same height and radius.
Thus, the volume of the hemisphere is the volume of the cylinder minus the volume of the cone, or &pir2(r) &minus 1/3&pir2(r). This gives us 2/3&pir3.
And of course, the volume of the hemisphere is exactly half that of the whole sphere, so we just multiply by 2 to get the volume of the whole sphere: 4/3&pir3.
Update: An explanation for how the volume of a cone is derived can be found starting here.
The big idea behind the demonstration is this: Suppose you have two different solid figures, both with the same height and the same width sitting side by side. Choose a height and make a horizontal slice at that height through both solid figures. If you can show that the resulting cross sections have the same area no matter where you make that slice, then the two solid figures MUST have the same volume.
We start with a cone whose height and radius are the same (left figure). We'll label both the height and the radius with b. Now imagine enlarging this cone proportionally to form the larger cone you see at the right. Its height and radius are similarly equal. We'll label the height and radius of the larger cone with r.
One very important idea to “see” before we move on. The area of the circular base of the smaller cone can be represented as &pib2. And no matter how we shrink or enlarge this smaller cone (proportionally), its base will ALWAYS have an area of &pib2.
Without stopping for a moment to think about that, it may sound ridiculous. Of course the circular base is always going to have an area of &pib2. The radius is, after all, b.
But not so fast. The height is also b, so what we’re saying is that no matter what the height of the smaller cone is, its circular base will ALWAYS have a radius equal to the height of the cone. Thus, the smaller cone will always have an area of &pib2. In other words, we know that the area of the base is a function of its radius (because we're using only positive values of b). But in this case, we've set it up so that the area of the base is a function of its height.
Okay, so let's take this bi-conical construction and put it inside a cylinder with the same radius and height as the larger cone.
Now let's make a horizontal slice all the way through the larger cone and the cylinder. We'll make our slice right at the circular base of the smaller cone.
If we took this cross section out of the cylinder, we would see a perfect circle, and that circle would have a radius of r. Thus, the circle would have an area of &pir2. In fact, since the radius of the cylinder remains the same up and down its height, we would see the same circle with the same area no matter where we made our slice. Thus, no matter where we make our slice, the resulting circular cross section would have an area of &pir2.
Now take a look at the cross section above. If we take that cross section and remove the area of the circular base of the smaller cone, what would the area of the cross section be?
Well, that's easy. Since the area of the circular base of the smaller cone is &pib2, we just subtract that area from the total area of the circular cross section. The result is &pir2 &minus &pib2.
Here's the kicker. We explained that no matter where we make the slice through the cylinder, the resulting circular cross section would have an area of &pir2. And we also explained that no matter where we make the slice through the cylinder, the area of the circular base of the cone would have an area of &pib2. THEREFORE, no matter where we make our slice, the resulting cross section without the area of the circular base of the cone will have an area of &pir2 &minus &pib2. Using the Distributive Property to rewrite this, we get &pi × (r2 &minus b2).
Okay, now let's construct a hemisphere (a half-sphere) with the same radius and height as the cylinder. We'll make a slice through this hemisphere at the same height (b) as we made our slice through the cylinder to form another circular cross section.
Now take a look at the right triangle formed in the hemisphere. The hypotenuse of this triangle is equal to r, because all the points that form a hemisphere are an equal distance from a fixed center point. And we see that one of the legs of this right triangle has a length of b. We can also see that the side length that we don't know is the radius of the circular cross section.
We can't say that this side length is r, because it's slightly smaller than that. As you can see, the "higher" we make our slice through the hemisphere, the smaller the resulting circular cross section would be. Similarly, the "lower" we make our slice, the larger the resulting cross section would be. Only at the base of the hemisphere would the "cross section" have a radius of r. Otherwise, it's going to be smaller than r.
So, what is this side length, this radius of the circular cross section in the hemisphere? Well, as it turns out, we don't need to find this side length. We can just figure out the square of this side length. Using the Pythagorean equation, we see that the square of the missing side length is equal to r2 &minus b2.
So since the area of the circular cross section of the hemisphere is equal to &pi × radius2, and radius2 in this case is r2 &minus b2, we can write the area of the cross section of the hemisphere as &pi × (r2 &minus b2).
This, of course, is the exact same area as the cross section in the cylinder at the same height with the area of the circular base of the smaller cone removed. And, just as it was for the cone inside the cylinder, the area of the circular cross section in the hemisphere is a function of its height. (The Pythagorean equation will not change. No matter where the cross section occurs, the radius will be "hypotenuse squared minus length of leg squared.") Thus, no matter where we slice through the hemisphere, the resulting cross section will ALWAYS have an area of &pi × (r2 &minus b2). Here are the two figures again.
Both the cylinder and the hemisphere have the same height and the same radius. And at every point along their heights, the circular cross sections have the same area so long as we remove the area of the base of the cone in the cylinder.
What this means is this: The volume of a hemisphere is equal to the volume of a cylinder with the same height and radius minus the volume of a cone with the same height and radius.
Thus, the volume of the hemisphere is the volume of the cylinder minus the volume of the cone, or &pir2(r) &minus 1/3&pir2(r). This gives us 2/3&pir3.
And of course, the volume of the hemisphere is exactly half that of the whole sphere, so we just multiply by 2 to get the volume of the whole sphere: 4/3&pir3.
Update: An explanation for how the volume of a cone is derived can be found starting here.
Labels: mathematics
Comments:
well. whattaya know.
this is news to me.
but where does the formula
V = \pi*r^2*h / 3
(for the volume of a cone)
come from? i can derive this
(and in fact, did, yesterday)
using calculus techniques,
but if we're going that route
we might as well skip this
(admittedly cool) ad hoc
construction and directly
calculate the volume of a sphere
by calculus and get it over with.
i'm reasonably sure there's some
more elementary way to see
that the volume of a cone
is what it actually is;
archimedes knew all this stuff.
i just figure you still owe us ...
I was hoping no one would ask me this. But, damnit, you're right.
And when you're right, you're right.
I'll start thinking about this one tomorrow.
Thanks for the great comment!
The cylinder proof is based on the pyramid proof. The pyramid proof, goes something like this. Start from a cube. If you can cut it into three identical rectangle pyramids, you have demonstrated that the volume of the pyramid is 1/3 that of the cube.
The pyramids have to be a rectangle pyramids, i.e. one with the vertex directly above one corner of the base, in the same fashion that a triangle rectangle does in 2D. Pick the bases from three adjacent faces of the cube (i.e. they all touch each other). The vertex point of the pyramids is opposite to the corner in common among the three bases. It's easy to build one out of paper or cardboard if you have difficulty visualizing how it goes in 3D.
The same turns true for a parallelepiped. You can extend this line of reasoning to a cone, as composed by many infinitesimal pyramids.
Giovanni Ciriani
Actually the rigorous demonstration works for a prism with a triangular base. And from there it can be extended to the cone.
Giovanni Ciriani
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