How do I show that the area of a circle is pi r^2 with integration?
you intefrate the circumference :)
integrate even lol
2pi r from 0 to radius of circle
integrate \[\sqrt{r ^{2}-x ^{2}}\] from 0 to r . Multiply by 4
what that amounts to is adding up all the circumferences from 0 to the radius whih = area of the circle
\[\int_{0}^{r}2\pi.r dr\] \[2\pi \int_{0}^{r}r.dr\] \[2\pi \frac{r^2}{2}-2\pi \frac{0}{2}=\pi r^2\]
Thank you soo much!!!!
youre welcome :) i accidently discovered that when I tried finding the volume of a solid the wrong way lol
Hi, Amistre I think what you want to say is \[\int_0^{2\pi}r\,dr\]
watchmath I integrated your formula, but I got 2 pi^2
Ah sorry, it should be double integral \(\int_0^{2\pi}\int_0^r r\,drd\theta\) But I guess you haven't learn double integral
not yet
I think the more traditional one is the following \[2\int_{-r}^r\sqrt{r^2-x^2}\,dx \] But to compute this integral you need to use the trigonometric substitution.
r is a constant right?
i typed it as i see it :) \[2\pi \int_{0}^{r}r.dr\] its the shell method for area instead of volume
area is the sum of all the circumferences of circles from 0 radius to full radius
top r might be better expreesed as x tho
I see that know amistre :D
\[\frac{2\pi r^2}{2}|_{0}^{x}\]
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