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

  • 1
    $\begingroup$ 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? $\endgroup$ Feb 5, 2019 at 2:56
  • 1
    $\begingroup$ @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. $\endgroup$ Feb 5, 2019 at 3:11
  • $\begingroup$ @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. $\endgroup$ Feb 5, 2019 at 3:52
  • $\begingroup$ 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. $\endgroup$
    – Moti
    Feb 5, 2019 at 5:46
  • $\begingroup$ 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. $\endgroup$
    – cr001
    Feb 5, 2019 at 7:03

1 Answer 1


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.

  • $\begingroup$ I'm unable to follow you, how does the distance each point travels even matters when at the end it makes a sphere? I understand that the error is due to counting the axis and the poles multiple times during the rotation. But if it is the only contribution to the error,then why isn't it consistent throughout all the similar figures? Like a cone. $\endgroup$ Feb 7, 2019 at 10:32
  • $\begingroup$ @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. $\endgroup$
    – fang
    Feb 7, 2019 at 20:06
  • $\begingroup$ assume that, you are indeed correct for the volume and area. But how would you explain the same logic for surface area? In this case all that travels is at the same distance from the center of the sphere $\endgroup$ Jun 2, 2020 at 18:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .