It's the end of semester and we are covering polar coordinates. We are answering questions of areas, slopes, orientation, and general shape. What troubles me is that it's the end of the semester and the instructor barely touched on the topic, heavily skipping subsections. I wondered if I should cover portion that he skipped or simply go with his flow. I always did everything that I was asked and went beyond the homework trying to explore some more. If skipping this section will mean crippling my later progress I will do it after the semester end or even grind my teeth and overcome fatigue.
In the future I will be taking multivariable calculus, linear algebra, time series, financial math, and a bunch of probability and stats ending on stochastic method of approximation. The end goal is to become an Actuary. What is your advice? Should I go an extra mile on this one or spend more time studying for Actuarial exams? Will polar coordinates occur at later classes?
Yes.
Since you aim to be an actuary, here's an example that is relevant to you. Consider the function $f(x) = e^{-x^2}$; you may already know that this is essentially the density curve of the normal distribution. (There are some constants missing, but those never make your life hard anyway, so let's ignore them.) Since this is the density of a very important random variable, the integral $$\int_{-\infty}^{\infty} e^{-x^2} \, \text{d} x$$ matters a lot, because we need to know what it is in order to properly scale that function to get it to actually be the density of a random variable.
When you set out to compute that integral, you might notice that you get stuck rather quickly. You won't be able to find a simple antiderivative of that function, and it's true (though not at all trivial) that one does not actually exist. So at face value, we feel stuck and that integral seems intractable... until we think of a clever trick.
First, we should note that the integral certainly converges; for instance, you can bound the integrand by $e^{-|x|}$. We'll call the value of the integral by $C$. Let's multiply two copies of that integral together: $$C^2 = \left( \int_{-\infty}^{\infty} e^{-x^2} \text{d} x \right) \left( \int_{-\infty}^{\infty} e^{-x^2} \text{d} x \right)$$ The variable of integration in either integrand is obviously unimportant, so let's change the second one to $y$ and smash them together with Fubini's Theorem: $$C^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} e^{-y^2} \, \text d y \, \text d x = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, \text d y \, \text d x$$ This still doesn't seem any better... until we switch to polar coordinates. We're integrating over the entire plane, which makes the limits of integration easy: $$C^2 = \int_{0}^{2 \pi} \int_{0}^{\infty} e^{-r^2}r \, \text d r \, \text d \theta$$ The switch to polar coordinates now makes the problem a chipshot with a $u$-substitution, from which you can show that $C = \sqrt{\pi}$. (I'll leave the details as an exercise.)
My point of this example is: you might think of polar coordinates as something esoteric, or something that's maybe useful only for physicists, but they pop up everywhere, including places that you don't expect. In this case, they deliver one of the most fundamental integral computations for much of probability and statistics.
So in short: polar coordinates are worth your time.