Well we work with spheres in derivative and integral problems

 

It's interesting to note that derivative of volume (wrt radius)

[math] V = \frac{4}{3}\pi r^3 [/math]

[math] V' = 4\pi r^2 [/math]

Is the formula for surface area.

 

And circles are the same:

[math] A = \pi r^2 [/math]

[math] A' = 2\pi r [/math]

Area goes to circumference.

 

Now I know this isn't just a coincidence. If someone could explain why this is that'd be great. And also, what happens if you found the integral of the Volume of the sphere (or the Area of a circle) ?

(they're [math] \int \frac{4}{3}\pi r^3 dr = \frac{\pi r^4}{3} [/math] and [math] \int \pi r^2 dr = \frac{\pi r^3}{3} [/math] ignoring the + C. I think thats what they'd be)

btw not very good with integrals yet so I'm not even sure if I did them right

Circumference of circle

 

[math]2\pi r[/math]

 

Infinitesimal area of circle

 

[math]2\pi rdr[/math]

 

Integrate over this between 0 and R (the radius of the circle)

 

[math]\int _0^R2\pi rdr=\pi R^2[/math]

 

Similarly with a sphere.