Finding the value of pi using calculus. I am new to calculus and not have much knowledge of how it works, I thought about finding the value of $\pi$ using the idea of dividing a circle into multiple parts and finding its area and thus extracting $\pi$ from it, the formula I got was 
$$\lim_{d \to 0} 4d\sum_{n=1}^{\lfloor 1/d\rfloor} \sqrt{1-(nd)^2} $$
Is there any way to simplify this formula using calculus as the only way currently I can solve this is taking $d$ as small value and adding each part manually.
Thanks for answers in advance
 A: $\pi$ can be found using calculus as the arc length of a semicircle of radius $1$ (this method is similar to the one you suggest...). Since the circle has equation $x^2+y^2=1$, the arc length formula gives $$\pi=\int_{-1}^1\frac{1}{\sqrt{1-x^2}} dx$$
A: There are many ways to approximate $\pi$, all reliant on limits if not calculus in some way or another. (Well, arguably all use calculus because they require a definition of $\pi$, which is either as an arc length from intrinsic calculus or an area from integral calculus.) You can write $\pi$ as various integrals and sums, but in practice we compute these integrals by cutting an area into $n$ strips (say), and so every method of approximating $\pi$ can be seen as defining a sequence $\{\pi_n\}_{n\in\Bbb N}$ such that $\lim_{n\to\infty}\pi_n=\pi$.
The most important difference between these options is how quickly $\pi_n$ approaches $\pi$ as $n$ grows. If you cut a circle into $n$ sectors each subtending $2\pi/n$ radians and approximate the sectors as triangles, the first-order approximation $\sin\theta\approx\theta$ of the area gives a $O(1/n^3)$ error per sector, and $O(1/n^2)$ for $\pi$ itself. Unfortunately, $|\pi_n-\pi|\in O(n^{-p})$ isn't as impressive as it sounds at first, because each extra decimal place of precision requires multiplying $n$ by $10^{1/p}$. It's not so bad if you can increase $p$ (which you can), but it'd be much better if the error shrank exponentially, so an extra decimal place only requires adding a constant to $n$.
Amazingly, many techniques can do even better than that. Exponentially shrinking errors $\epsilon_n:=|\pi_n-\pi|$ satisfy $\epsilon_{n+1}\approx r\epsilon_n$ with $0<r<1$. But if $\epsilon_{n+1}\in O(\epsilon_{n+1}^p)$ with $p>1$, we have a form of superlinear convergence in which the number of correct decimal places multiplies approximately by $p$ every time you increase $n$. It's not important to make $p$ much bigger than $1$, and approaches achieving $p>2$ tend to be (i) so computationally complex at each iteration they're not necessarily "faster" overall and (ii) dependent on much more advanced theory. For your purposes, $p=2$ may be a sweet spot, e.g. if you use this approach.
A further benefit of the above link is that it allows you to build on previous calculations; a sequence of the form $\pi_{n+1}=f(\pi_n)$, or at least one of the form $\pi_n=g(a_n^{(1)},\,\cdots,\,a_n^{(k)}),\,a_{n+1}^{(i)}=f_i(a_n^{(i)})$, allows you to continue if you want more precision, whereas if you didn't chop up a circle you have to start over again with more slices. Or do you? Your original approach can be fruitful if you keep doubling the number of slices. A calculation with a $3\times 2^{28}$-gon obtained the equivalent of $17$ decimal places in 1424.
