If you want to use integration, wouldn't it be easier to start from the fact that the circumference of a circle, radius t, is 2πt by definition. Then the area of a circle, radius r, is simply the integral of 2πt.dt from t=0 to t=r. That turns out to be [2πt^2 / 2] evaluated at r - 0 and that is πr^2 - π0^2 = πr^2.
If you want to use integration, wouldn't it be easier to start from the fact that the circumference of a circle, radius t, is 2πt by definition.
Then the area of a circle, radius r, is simply the integral of 2πt.dt from t=0 to t=r.
That turns out to be [2πt^2 / 2] evaluated at r - 0 and that is πr^2 - π0^2 = πr^2.
Yep it would. There are a number of ways of doing it. This is just one which I found interesting, hence the video. Thanks for your comment though.