CanadaAotS Posted January 11, 2006 Share 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) Link to comment Share on other sites More sharing options...
CanadaAotS Posted January 11, 2006 Author Share Posted January 11, 2006 btw not very good with integrals yet so I'm not even sure if I did them right Link to comment Share on other sites More sharing options...
□h=-16πT Posted January 12, 2006 Share 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. Link to comment Share on other sites More sharing options...
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