Line Integral $\int_{\gamma} \frac{y}{\sqrt{x^2+y^2}}ds \qquad \qquad \gamma(t)\colon t\mapsto(\cos^3t,3\cos^2t\sin t)$ as $t \in (a,b)$

Question

• Hey there, I'm asked to find $$\int_{\gamma} \frac{y}{\sqrt{x^2+y^2}}ds \qquad \qquad \gamma(t)\colon t\mapsto(\cos^3t,3\cos^2t\sin t)$$ as $t \in (a,b)$. As we have regularity both in the field and in the curve, we can easily apply the standard formula: $$\int_{\gamma} \frac{y}{\sqrt{x^2+y^2}}ds = \int_a^bF(\gamma(t)) ||\gamma'(t)||dt$$ but this leads to terrible computations. Is there anything smart I can do?

• Can I do something smart to simplify my operations when I'm going (like in this case) over a curve which is of the kind $t \mapsto (f(t),kf'(t))$ where $k$ is a constant?

Answer 1

You are asked to integrate $\frac{y}{\sqrt{x^2+y^2}}$ over a sort of lemniscate which goes through the first, second, third and fourth quadrant in this exact order. Is there a reason for expecting surprising cancellations/simplifications when $a$ and $b$ are generic?

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If you do not get scared by actually performing the substitutions, assuming $a,b\in[-\pi/2,\pi/2]$ the integral becomes

$$ -\int_{\cos(a)}^{\cos(b)}\frac{x^3}{\sqrt{9 x^2 \left(8 x^4-11 x^2+4\right)}}\sqrt{9x^4-8x^6}\,dx$$ or $$ -\frac{1}{3}\int_{\cos(a)}^{\cos(b)}\frac{x^4\sqrt{9-8x^2}}{\sqrt{8 x^4-11 x^2+4}}\,dx=-\frac{1}{6}\int_{\cos^2(a)}^{\cos^2(b)}\frac{x^2\sqrt{9-8x}}{\sqrt{x(8x^2-11x+4)}}\,dx$$ which is an elliptic integral. So there is no simple expression for general values of $a$ and $b$.

Answer 2

Notice that the curve is $2\pi$-periodic and the symmetry the curve has across the $x$ axis (since $y(t)$ is an odd function-this is different from the integrand being odd, which it is separately). Thus we have that

$$\int_\gamma\frac{y}{\sqrt{x^2+y^2}}\:ds = 0$$

by odd symmetry.