# Help with deriving surface area of a sphere

Basically, I was playing around trying to derive the surface area of a sphere.

My logic is, looking at one half of a sphere, if we slice the circle out of the very middle it will be the biggest circle and have a circumference of $$2(\pi)(R-0)$$. And if we keep slicing and "unroll" the circumferences into a line they will get just a tiny bit shorter each time. Do so with all the slices and stacking the lines that get shorter and shorter approaching a length of $$0$$, these resemble the shape of a triangle. We can get a linear function for this triangle and then integrate it to add up all the lines. (the area of the triangle) We would have to double our final result to get the other half of the sphere.

The solution I came to was $$6(\pi)r^2$$ but that is wrong : I see it is actually $$4(\pi)r^2$$.

The disc-repancy (pun intended) is suspiciously equal to subtracting 2 circles with rad R (the center of our sphere) away from my result. Can anyone help me identify flaws in my logic or math, I am trying to improve my math and integration skills.

my notes, I hope somewhat legible

Thank you and best.

• You should try to remove the handwritten notes & try to type some more calculations here. While I have upvoted , that unnecessary image might attract downvotes later. Draw with Computer/Software & then type the calculations with MathJAX will be the Best !
– Prem
Commented Apr 28 at 8:27
• Here's a non-calculus proof: youtube.com/watch?v=GNcFjFmqEc8
– Nate
Commented Apr 28 at 22:47
• That was a nice video , @Nate , though at around 10:00 , we hear something like this "this kind of reasoning is essentially calculus without the jargon" : Still , the Idea was Interesting to me !!
– Prem
Commented Apr 29 at 6:56

When you try taking the thin slices without considering the Sphere Curvature & the $$\theta$$ which controls the radius of the slices , you will get something like a cone : Naturally , you will get wrong answer.

When you are taking the thin slices while considering the Sphere Curvature & the $$\theta$$ which controls the radius of that slice , you will get the Semi-sphere : Ideally , you should get the right answer.

At angle $$\theta$$ , we have the Purple line which is the radius of sphere $$r$$ , while the Green line is $$r \sin \theta$$ & the Blue line is twice $$r \cos \theta$$
Hence the thin slice has radius $$r \cos \theta$$ & thickness $$r d\theta$$ (due to Sphere Curvature) & hence slice area is $$2 \pi \times r \cos \theta \times r d \theta = 2 \pi r^2 \cos \theta d \theta$$

Sphere Surface Area is $$2 \times \int_0^{\pi/2} 2 \pi r^2 \cos \theta d \theta = 4 \pi r^2 [\sin \theta]_0^{\pi/2} = 4 \pi r^2$$

Same logic will give Sphere Volume too.

Here is a Zoom-In for Cone & Sphere :

We can see that the Blue thin slice has thickness shown with Green tiny Piece.

Cone & Sphere have Curvature to make a Circle which we can unroll with the Blue lines.
Checking the Green lines , we see that Cone has no change in slant , while Sphere has varying slant , according to $$\cos \theta$$.
That makes the calculation given here valid.

• Thank you for the comment. But, I think my curiosity persists. Am I not accounting for the curve of the sphere by taking tinier and tinier slices, and the curve of the slices of circles by unrolling the circumference? Intuitively, I feel that all the information of the surface area would be encapsulated in that logic? Commented Apr 28 at 13:47
• (1) Check Surface Area of Cone with your logic & see whether what you get is right. If yes , then that logic should not be valid for Sphere , right ? Something should change , right ? (2) Each slice is curved in 2 Dimensions for Sphere : that must be handled. (3) Unrolling handles 1 Dimension. Trigonometry handles the other Dimension.
– Prem
Commented Apr 28 at 13:59
• Okay, I will check with a cone. But I feel that would be incorrect? Like the second dimension you are referring to, wouldn't that curve be considered from taking smaller and smaller slices? Like if I have the biggest middle slice, and move over and infinitely small amount, that smaller circle, that would be that 2nd dimension you are talking about, no? (Also, I suck at traditional typesetting, but I tried to clean up my notes, I came a different answer but am mainly concerned with the thinking behind it.) Commented Apr 28 at 14:17
• ADDENDUM Image should make things a little more clear.
– Prem
Commented Apr 28 at 15:02