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 := $$A = 1/2rv\int_t^T dt = 1/2 rvT = 1/2 rv 2πr/v = πr^2$$

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

1 Answer 1


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!


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