I don't understand Stokes' Theorem. Every intuitive explanation feels so dumbed down that it feels like it says nothing, every rigorous and satisfying (I'm guessing) explanation seems intimidating, and every rigorous accessible explanation feels unsatisfying. I couldn't even begin to try to prove it to save my life.
$$\int_{\partial R} \omega = \int_R d\omega$$Despite this it's not that difficult to use for actual calculations, at least in the direction of going from integrating $\omega$ to integrating $d\omega$. Exterior differentiation is computationally a pretty easy process. I do, however, consistently struggle with doing the parts that should be easy like not screwing up basic fraction arithmetic or setting up the bounds of integration.
Given that I hope to use and teach Stokes' theorem in Multivariable Calculus But Good as the only theorem for integration, I figured it's important to practice doing computations with Stokes' theorem directly to make sure I can do them consistently correctly, as it would help me in teaching it while also perhaps helping build intuition that can help me understand the theorem and feel intuitively why it works and why its true.
Interestingly enough there don't seem to be that many examples of Stokes' theorem in use to evaluate integrals out in the wild if you look for practice just by a routine search on a search engine, so hopefully this can become a resource for people who want that too.
I stole the problems below from Paul's Notes.
Integrate the form $\omega = -yz\,dx + (4y+1)\,dy + xy\,dz$ over the circle of radius 3 at $y = 4$ and perpendicular to the y-axis with a counterclockwise rotation looking at the positive y-axis from the origin.
Solution: This is a 1-form integrated over a curve. Thus, the integral of interest is $\int_{\partial R} \omega$. We can apply the fundamental theorem to compute $\int_R d\omega$ instead.
$$\begin{align*} d\omega &= d(-yz)\wedge dx + d(4y+1)\wedge dy + d(xy) \wedge dz\\ &= (-y\,dz - z\,dy)\wedge dx + 4\,dy \wedge dy + (x\,dy + y\,dx) \wedge dz\\ &= -y\,dz \wedge dx - z\,dy \wedge dx + x \,dy \wedge dz + y\,dx \wedge dz\\ &= 2y\,dx\wedge dz - z\,dy\wedge dx + x\,dy\wedge dz \end{align*}$$The region $R$ within the circle $\partial R$ is given by the function $y(x,z)=4$, provided we restrict the domain to $-3 \le z \le 3$ and $-\sqrt{9-z^2} \le x \le \sqrt{9-z^2}$. In this situation, $dy = 0$, so $d\omega = 8\,dx \wedge dz$, where we put $dx$ before $dz$ since seen from the xz plane with the positive parts of the axes in the usual place, the circle is counterclockwise so this captures the orientation, and:
$$\begin{align*} \int_R d\omega = \int_R 8\,dx\wedge dz = \int_{-3}^3\int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}} 8\,dx\,dz = 16\int_{-3}^3 \sqrt{9-z^2}\,dz = 72\pi \end{align*}$$I got this one completely right first time purely by hand which is hype.
Verification: We can parametrize the circle as $\varphi(\theta) = (3\cos(\theta), 4, 3\sin(\theta))$. Then we have (forgive the potentially dubious notation) that:
$$\begin{align*} \omega_\varphi \circ d\varphi &= -4\cdot3\sin(\theta)\cdot d(3\cos(\theta)) + 3\cos(\theta)\cdot 4\cdot d(3\sin(\theta))\\ &= -36\sin(\theta)\,d(\cos(\theta)) + 36\cos(\theta)\,d(\sin(\theta))\\ &= (36\sin(\theta)^2 + 36\cos(\theta)^2)d\theta = 36\,d\theta \end{align*}$$Therefore the integral of interest is:
$$\int_{\partial R} \omega = \int_{-\pi}^{\pi} 36\,d\theta = 2\pi\cdot36 = 72\pi$$Which is way easier to be honest.
Integrate the form $\omega = (3yx^2 + z^3)\,dx + y^2\,dy + 4yx^2\,dz$ over the triangle curve with vertices $(0, 0, 3), (0, 2, 0), (4, 0, 0)$ with counterclockwise rotation looking down on the xy plane.
Solution: This is a 1-form integrated over a curve. Thus, the integral of interest is $\int_{\partial R} \omega$. We can apply the fundamental theorem to compute $\int_R d\omega$ instead.
$$\begin{align*} d\omega &= d(3yx^2 + z^3)\wedge dx + d(y^2)\wedge dy + d(4yx^2) \wedge dz\\ &= (3x^2\,dy+3xy\,dx+3z^2\,dz)\wedge dx + (4x^2\,dy + 8xy\,dx) \wedge dz\\ &= 3x^2\,dy\wedge dx + 3z^2\,dz\wedge dx + 4x^2\,dy\wedge dz + 8zy\,dx\wedge dz\\ &= 3x^2\,dy\wedge dx + 4x^2\,dy\wedge dz + (8zy - 3z^2)\,dx\wedge dz \end{align*}$$The region $R$ within the circle $\partial R$ is given by the function $z(x,y)=3-3y/2-3x/4$, provided we restrict the domain to $0 \le y \le 2$ and $0 \le x \le 4-2y$. In this situation, $dz = -3\,dy/2 - 3\,dx/4$, and furthermore:
So $d\omega = (9z^2/2-12xy)\,dx\wedge dy$. We put $dx$ before $dy$ because looking top down onto the xy plane the triangle goes counterclockwise and this captures the orientation. Now:
$$\int_R d\omega = \int_0^2\int_0^{4-2y} \left(\frac{9}{2}\left(3-\frac{3y}{2}-\frac{3x}{4}\right)^2 - 12xy\right)dx\,dy = -5$$Full disclosure: I had to set up the domain restrictions a bunch of times and do the final double integral with WolframAlpha.