Using calculus to get the surface area of a sphere

[hidden dagger]

When I use the same method to find the volume (using area instead of circumference), I got the correct formula. Can somebody tell me why it works for volume but not surface area?

[chilly karma]

pretty sure that if you fix $f:[0,\infty)\to[0,\infty),f(x)=\sqrt{1-x^{2}}$ you need a specific formula regarding surface area of a solid of revolution

[split marsh]

chudcel

[chilly karma]

which returns the formula you want

or more generally $f(x)=\sqrt{r^{2}-x^{2}}$

[split marsh]

chudcel

[chilly karma]

i did the integral once but you actually have to tweak it alot for it to return this

[upper hornet]

Real ones differentiate the volume to get the surface area

[summer wing]

This is easier to intuitively see in polar co-ordinates. The area element of a unit sphere $\dd{S} = \dd{\varphi} \dd{\theta} \sin{\theta}$. So far so good. Here $\theta \in [0, \pi]$ and $\varphi \in [0, 2\pi]$. Integrating over this domain is intuitively rotating a unit semicircle for a full revolution, which gives us the total surface area. What you have done instead is consider $\dd{\theta} \sin{\theta} = y \dd{x}$, where $y = \sqrt{1-x^2}$. This is of course incorrect as $y \dd{x} = \sin{\theta} \dd{\cos{\theta}}= - \sin^2{\theta} \dd{\theta}$.

[split marsh]

Killuminati

[hidden dagger]

oh ok

[hidden dagger]

thank you

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