# Finding area of a circle from circular motion

We can find the change in area by $$dA=1/2 |r⃗ ×dr⃗|$$ $$dA = 1/2 r dr \sin(θ) = 1/2 r dr$$ since velocity vector is orthogonal to r

$$dA = 1/2 r vdt$$ T =2πr/v $$A = 1/2\int_t^T r vdt$$

couldn't go further, how can I find the area with little time stamps' areas added to become the whole?

Edit: Nevermind, it works, just when I edited it Bela Bahaa also published it := $$A = 1/2rv\int_t^T dt = 1/2 rvT = 1/2 rv 2πr/v = πr^2$$

• Hi, welcome to Math SE. Hint: $\int_0^Tcdt=cT$ if $c$ doesn't depend on $t$.
– J.G.
Feb 4 at 10:16

The steps mentioned here are in fact correct. You can continue by noticing that $$r$$ is the radius of the circle, which is a constant. Thus, you get that $$A = \frac{r}{2} \int_t^T v dt$$

Notice now that we can replace the $$v dt$$ with $$dr$$, and we change the limits of the integration to $$0$$ and $$2 \pi r$$ since we are no longer integrating with respect to time but rather the displacement covered.

So, the final integral is

$$A = \frac{r}{2} \int_0^ {2\pi r} dr$$

You can easily notice that this yields

$$A = \frac{r}{2} \times 2\pi r$$

So, we arrive to the final solution which is

$$A = \pi r^2$$

Hope that helps!