Yesterday I was asked by a former student of mine (a very bright twelfth grader) to show him solutions to two problems that he found in an Analysis book, both on surface integrals. The problems rather look similar, and I managed to solve the first, however, I was stuck on the second one.
The first: Calculate
$$\int \frac {y dx + x dy} {x^2 + y^2}$$ over a segment of the straight line $y = x$ from $x = 1$ to $x = 2$. I solved this one: $$\int \frac {y dx + x dy} {x^2 + y^2} = \int_{1}^{2} \frac {2x dx} {2 x^2} = \int_{1}^{2} \frac {dx} {x} = \log 2.$$
The second: Calculate
$$\int x dy - y dx$$ over the curve $x = a (t - \sin t)$, $y = a (1 - \cos t)$. I am stuck on this one. What I've done so far is, substitute $dx = a (1 - \cos t) dt = y dt$, $dy = a \sin t dt$ then $$ x dy - y dx = a^2 (t \sin t - 2 + 2 \cos t) dt $$ and $$\int x dy - y dx = a^2 \int (t \sin t - 2 + 2 \cos t) dt.$$ But this is seemingly a wrong approach because it leads nowhere. I'm afraid I'll be embarrassed if I'll have to tell him I've been unable to solve it.
Note that
$$ \frac{1}{2}\oint_C xdy -y dx $$
is a particular case of the Green theorem that gives you the area of the surface bounded by $C$. And by the Green theorem again, the following equalities hold $$ \frac{1}{2}\oint_C xdy -y dx =\oint_C xdy =-\oint_C y dx, $$
so if you are having trouble with your integral, you can try working with the two other ones. This is one way of getting rid of the term that you cannot integrate, although integrating by parts should easily work.
In your case your result will give you twice the area beneath the curve, and above the $x$ axis.
Note. Although your curve is not closed, you can close it with a line segment on the $x$ axis, and integrating on this segment will give you $0$ (because $y=dy=0$ on this line), so you can consider that the curve is closed (provided you are working with $t$ in $[0,2\pi]$).