Green's Theorem to calculate a line integral

Question

Use Green's Theorem to evaluate the line integral $\int_C y^3dx + x^3dy$ where $C$ is the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$, so I did this using symmetry. However, I friend switched to polar coordinates, and found this instead. We have no way of knowing if what we did is correct, so we wanted to verify.

Answer 1

Anyway, regardless of what the constants are, here's a general trick for integrating over ellipses. Once you perform Stokes' theorem in this example, you'll end up with an integral like \begin{align} I&= \int_E (c_1 x^2 + c_2 y^2)\, dx \, dy \end{align} where $c_1, c_2 \in \Bbb{R}$ are constants, and $E$ is the region bounded by the ellipse described by $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}\leq1$. The first trick to notice is that integrals over ellipses can be transformed to an integral over the unit disk by a simple linear change of variables: \begin{align} \begin{cases} x &= au \\ y &= bv \end{cases} \end{align} where $u^2 + v^2 \leq 1$. It is easily seen that this maps the unit disk, $D$, onto the ellipse $E$. Hence, we get \begin{align} I &= \int_D (c_1a^2u^2 + c_2 b^2 v^2) \, (ab) \, du \, dv \end{align} Now, by symmetry, one can argue that \begin{align} \int_D u^2 \, du \, dv = \int_D v^2 \, du \, dv &= \dfrac{1}{2}\int_D (u^2 + v^2) \, du \, dv \\ &= \dfrac{1}{2}\int_0^1\int_0^{2\pi} r^2 \cdot r \, d\theta \, dr \\ &= \dfrac{\pi}{4}, \end{align} where in the second line, I changed to polar coordinates. If you plug this into the above expression for $I$, you'll find that \begin{align} I &= (c_1a^2 + c_2 b^2)(ab) \left( \dfrac{\pi}{4}\right). \end{align} (Of course, don't ever try to memorize this result, keep in mind how it was derived, and then if need be, repeat it in the future).

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In your particular case, we have $c_1 = 3, c_2 = -3$ and $a=3, b=5$, which yields $I = -180 \pi$. If instead you meant to type the line integral $\displaystyle\int_C y^3 \, dx - x^3 \, dy$, then we have $c_1=c_2 = -3$ and $a=3, b=5$, which yields $I = -\dfrac{765 \pi}{2} $, which agrees with your result.