CanadaAotS Posted January 11, 2006 Posted January 11, 2006 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)
CanadaAotS Posted January 11, 2006 Author Posted January 11, 2006 btw not very good with integrals yet so I'm not even sure if I did them right
□h=-16πT Posted January 12, 2006 Posted January 12, 2006 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.
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