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Let $C$ be the curve parametrized by the equation $r(t) = 2\cos^3(t) i + 2\sin^3(t) j$ for $t \in [0,2\pi]$. I want to find the line integral

$$ \oint_C y^2 \,dx + x \,dy . $$

I evaluated it directly and the answer appears to be $\frac{3\pi}{2}$. But I'm supposed to use Green's Theorem and I don't know how to set up the integral. If the region enclosed by $C$ is $D$, then the integral is

$$ \iint\limits_D 1 + 2y \, dy\, dx $$

But I have no idea how to find the region $D$. I tried to set $x = 2 \cos^3(t)$, which implies $ y \in [-k,k] $ for

$$ k = 2(1- (\frac{x}{2})^{2/3})^{3/2} $$

So the integral would become

$$ \int_{-2}^2 \int_{-2(1- (\frac{x}{2})^{2/3})^{3/2}}^{2(1- (\frac{x}{2})^{2/3})^{3/2}} 1 - 2y\, dy\, dx .$$

But this is too messy to evaluate by hand.

So, how do I evaluate the line integral using Green's Theorem?

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Note that by symmetry $$I:=\int_{-2}^2 \int_{-2(1- (\frac{x}{2})^{2/3})^{3/2}}^{2(1- (\frac{x}{2})^{2/3})^{3/2}} (1 - 2y)\ dy dx=4\int_{0}^2 \int_{0}^{2(1- (x/2)^{2/3})^{3/2}} 1dy dx\\ =8\int_{0}^2 (1- (x/2)^{2/3})^{3/2} dx=48\int_{0}^{\pi/2} \cos^4(t) \sin^2(t) dt\\ =48\int_{0}^{\pi/2} \cos^4(t)dt-48\int_{0}^{\pi/2} \cos^6(t)dt$$ where we let $x=(2\sin(t))^3$. Now, for $n\geq 1$, by integration by parts, $$\int_{0}^{\pi/2} \cos^{2n}(t)dt=\frac{2n-1}{2n}\int_{0}^{\pi/2} \cos^{2n-2}(t)dt$$ which implies that $I=3\pi/2$.

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  • $\begingroup$ Yes, but this just reduces to integrating $(1-x^{2/3})^{3/2}$, which is also tricky...unless I'm missing something. $\endgroup$ – aras Jun 30 '17 at 11:07
  • $\begingroup$ OK, thank you for explicitly showing me the substitution $x = (2\sin(t))^3$. I see why this solution works. However, isn't this essentially just reducing the computation back to the line integral? If I compute $\int_{0}^{2\pi} \langle y^2, x\rangle \dot \langle x'(t), y'(t) \rangle dt $ I get $$ \int_{0}^{2\pi} -24 \sin^7(t) \cos^2(t) + 12 \sin^2(t) \cos^4(t) $$ The $\sin^7 \cos^2$ term integrates to $0$, and the $12 \sin^2 \cos^4$ term simplifies to $$ 12 \int_{0}^{2\pi} \cos^4(t) - \cos^6(t) dt $$ which is what you evaluated in your answer. $\endgroup$ – aras Jun 30 '17 at 11:22
  • $\begingroup$ @aras Yes, so are able to find the final result? $\endgroup$ – Robert Z Jun 30 '17 at 11:24
  • $\begingroup$ Yes, but I was wondering if there was a cleaner way to apply Green's Theorem that avoided the computational mess of explicitly computing the line integral. $\endgroup$ – aras Jun 30 '17 at 11:25
  • $\begingroup$ Finally you have to evaluate the area inside the curve. I don't know if here you can avoid the computation of $\int_{0}^{2\pi} \cos^4(t) - \cos^6(t) dt$. $\endgroup$ – Robert Z Jun 30 '17 at 11:35
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Note that at any point $x$, we have $$y = 2\left(1-(x/2)^{2/3}\right)^{3/2}$$ so it sounds like you have the equation $$\left(\frac{y}{2}\right)^{2/3}+ \left(\frac{x}{2}\right)^{2/3} = 1$$

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  • $\begingroup$ How does this help me find the limits of integration of $D$? $\endgroup$ – aras Jun 30 '17 at 11:09
  • $\begingroup$ @aras since you have the description of the boundary, you can solve for whatever you like -- I even solved for $y$ for you... $\endgroup$ – gt6989b Jun 30 '17 at 14:46

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