When I imagine a circle($x^2+y^2=1$) and rotate it around the x axis, I get a shape of sphere.

So my question is why $(2πr)\timesπ$ and $(πr^2)\timesπ$ do not represent the area and volume of sphere.

I see that there are overlapped points especially in case of the volume, but still resulting values are too far away from the values obtained using spherical coordinates system.

What am I missing?

  • $\begingroup$ For the surface area of the sphere, consider a point along the perimeter of the original circle. The distance that it travels around the axis of rotation depends on where the point originally is. It could be that it doesn't move at all during the rotation, or it could make a large circle. As you can intuitively see from this, you need to use integration to derive the surface area of the sphere. $\endgroup$
    – Matti P.
    Jul 13, 2018 at 6:24
  • $\begingroup$ By the way, if you want to make a sphere from $x^2 + y^2 = 1$, maybe you want to rotate around the $z$-axis, no? Otherwise it doesn't change at all. $\endgroup$
    – Matti P.
    Jul 13, 2018 at 6:25

1 Answer 1


In this way we have some overlapping which cause an overestimation of the volume. The correct method to use this idea is based on Pappus's Centroid Theorem for the surface and volume

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  • 1
    $\begingroup$ Ah!!!!Wonderful as usual!!!!Greetings. $\endgroup$ Jul 13, 2018 at 6:43
  • $\begingroup$ @PeterSzilas Credits to Pappus, a very clever boy! :) $\endgroup$
    – user
    Jul 13, 2018 at 6:47
  • $\begingroup$ en.m.wikipedia.org/wiki/Pappus_of_Alexandria .Greetings. $\endgroup$ Jul 13, 2018 at 6:54
  • $\begingroup$ In case of the surface area of sphere, I think (2πr)×π is the overestimated value as you also mentioned about V. Correct value would be 4πr^2. Problem is, 4πr^2 is greater than overstimated value (2πr)×π when r=5. Why is it so? $\endgroup$
    – NK Yu
    Jul 15, 2018 at 12:03

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