Volume and surface area of sphere, cone, cylinder etc

Why isn't the volume of a sphere: $$\pi^\text{2}r^\text{3}$$, instead it is $$\frac{4}{3}\pir^\text{3}$$? Like wise the surface area is 4$$\pir^\text{2}$$and not 2$$\pi^\text{2}r^\text{2}$$.

Simply take a 2D circle and rotate on same center and radius perpendicular circle and we get a sphere. But this isn't consistent among all the shapes which have a common axis.

I believe the only repeated/common things in this derivation are the pole of intersection and the axis.

Thanks for your help.

• I'm not clear on exactly what you're asking for. Why should the volume of the sphere be $\pi^2 r^3$ or the surface area of a circle be $2\pi r^2$ (note the actual value is not $4\pi r^2$ but rather is $\pi r^2$)? If you wish to see proofs of the proper formulas, then you can get this from the Sphere and Area of a circle Wikipedia pages. I believe you are asking about objects having a common axis, but why should different shapes have the same area or volume, even with a common axis? – John Omielan Feb 5 at 2:56
• @JohnOmielan 1)Im Talking about sphere. A sphere ,as per standard books and sources, has a volume of {(4/3)*pi*(r^3)}. A surface area of {4*pi*(r^2)}.2)The problem is if we do a different type of derivation like, rotating two 2D circles orthogonally, we get different answers. And the rest is in the question. – user163416 Feb 5 at 3:11
• @user163416 Thanks for your response. I'm sorry for misreading your question to think you were talking about the surface area of a circle instead of a sphere. As for getting different answers depending on your derivation, I believe you need to check exactly what volume or surface area you are determining to see why they don't match the correct ones for a sphere. – John Omielan Feb 5 at 3:52
• You need to develop the full "integration" to get the right result. It is clear that the surface of a sphere is larger than 2pir^2 - 1/2 the sphere is larger than pir^2. – Moti Feb 5 at 5:46
• You cannot just rotate a 2D circle to form a sphere, because the density is denser near the center than farther from the center. Recall the process of deriving the circle's area: We don't just rotate a line, instead we rotate a slim "triangle" and the derivation is based on the base times height divided by 2. Note this is why the "2" disappears from the "$2\pi r$" circumstance to become $\pi r^2$. For sphere case, things become much more complicate, if you want to do rotation, you need to rotate a "thin slice of orange" instead of a pure 2D circle. – cr001 Feb 5 at 7:03

Although a sphere can be formed by rotating a 2d full circle by 180 degree, during the rotation every point on the 2d circle will travel by a different amount of distance. For example, the two points that is farthest from the rotation axis will travel by a distance of $$\pi r$$, the other points will travel by a distance $$\pi d$$ (where $$d$$ is the perpendicular distance to the rotation axis and $$d < r$$) and the two points at the intersection of the 2d circle and the rotation axis will not moved at all (i.e., $$d=0$$). Therefore, you cannot simply multiply full circle's area $$\pi r^2$$ and circumference length $$2\pi r$$ by $$\pi r$$ to obtain sphere's volume and surface area.
• @OP: When rotating a line of length $L$ wrt a parallel axis to form a cylindrical surface, the surface area can be computed as $2 \pi rL$. This is because every point on that line will travel by the same distance. This is not the same as in the case of rotating a circle to form a sphere. – fang Feb 7 at 20:06