Calculating line Integral with polar coordinates

Question

I want to calculate the following line integral using polar coordinates and verify my calculation is correct. $$\oint\limits_{C}2x+y^2ds\, \\C=\{(x,y)\in \Bbb{R}^2 | x^2+3y^2=8\}$$ My conversion to polar coordinates is the following $$\gamma: [0,2\pi) \rightarrow \Bbb{R}^2 \\\gamma(t) = \begin{pmatrix}\sqrt8\cos(t)\\\ \frac{\sqrt8\sin(t)}{\sqrt3} \end{pmatrix}$$ Now it should be $$\oint\limits_{C}2x+y^2ds\, = \int\limits_{0}^{2\pi}4\sqrt2\cos(t)+ \frac{8\sin^2(t)}{3} dt = \frac{8\pi}{3}$$ Is this correct?

Answer 1

We need to evaluate $ \displaystyle \oint\limits_{C} (2x+y^2) \ ds $, where $C=\{(x,y)\in \Bbb{R}^2 | x^2+3y^2=8 \}$

While you got the correct answer based on advice by Jochen in comments, there is a bit of simplification that you can do. Notice that $f(x) = x$ is an odd function and as $C$ has symmetry about y-axis, its integral over $C$ will be zero. So you are left with evaluating $ \displaystyle \oint\limits_{C} y^2 \ ds \ $.

As per parametrization you used, $\gamma: [0,2\pi) \rightarrow \Bbb{R}^2, \left(\sqrt8\cos(t), \frac{\sqrt8\sin(t)}{\sqrt3}\right)$

$\gamma'(t) = \left(-\sqrt8\sin(t), \frac{\sqrt8\cos(t)}{\sqrt3} \right), ||\gamma'(t)|| = \sqrt{ \cfrac {8+ 16 \sin^2(t)}{3}}$

So, $ \displaystyle \oint\limits_{C} y^2 \ ds = \frac{16 \sqrt2}{3 \sqrt3} \int\limits_{0}^{2\pi} \sin^2(t) \sqrt{1 + 2 \sin^2(t)} \ dt \ $, which is an elliptic integral. Using online calculator, the answer is $ \approx 21.506$ as mentioned in your comments.