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I thought of deriving the formula for the surface area of a sphere using integration.

So, below are my calculations:- S.A. of sphere = 2 × S.A. of a hemisphere

Now thinking of each of the hemisphere as being made up of an infinite number of circles having their radii in the range r= R to r=0 and knowing that circumference of a circle = $2 \pi r$, we have$:$ $$S.A. = 2× \int_R^0 2 \pi r \,dr$$ $= 2× ( \pi R^2 - 0) $ $= 2\pi R^2$

Why is my derivation wrong?

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  • $\begingroup$ In this way you are finding the projection area of the hemisphere, which is of course, $\pi R^2$. $\endgroup$
    – velut luna
    Commented Oct 18, 2017 at 11:19
  • $\begingroup$ @velutluna, can you tell me why isn't my 'proof' doing what I intended it to do? What is wrong in my reasoning? $\endgroup$
    – HeWhoMustBeNamed
    Commented Oct 18, 2017 at 11:23
  • $\begingroup$ @samjoe, so what do I need to do basically? $\endgroup$
    – HeWhoMustBeNamed
    Commented Oct 18, 2017 at 11:25
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    $\begingroup$ You can cutting it into ribbons. But the width of the ribbon is not $dr$. $\endgroup$
    – velut luna
    Commented Oct 18, 2017 at 11:25
  • $\begingroup$ @MrReality I recommend you to first learn proper usage of infinitesimals. Here, you first need to find a small general area element on hemisphere. $\endgroup$
    – jonsno
    Commented Oct 18, 2017 at 11:28

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You need infinitesimal "ribbon" areas. Now it's ok to construct these areas as the circumfence $2 \pi r$ times their height d$h$. Note that this height is not d$r$ (then the ribbon would lie flat - that's why you get two times the area of a circle with your integration). Since the ribbon is slanted, a simple drawing of two similar triangles shows that

$$ \frac{d h }{d r}= \frac{R}{\sqrt{R^2 - r^2}} $$

So the total surface area, as you already started, is

$$S.A. = 2× \int_{h(r=R)}^{h(r=0)} 2 \pi r \,dh = 2× \int_R^0 2 \pi r \frac{R}{\sqrt{R^2 - r^2}} dr $$

Now you can integrate this, using the substitution $r = R \sin \phi$:

$$S.A. = 2× \int_0^{\pi/2} 2 \pi R^2 \sin \phi \, d \phi = 4 \pi R^2 $$

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