Evaluate the line integral $\int_cy^4\ dx+2xy^3\ dy$ where $C$ is the ellipse $x^2+2y^2=2$.
My attempt:
First, I need Green's Theorem:
$\int_cP\ dx+Q\ dy = \int\int_D\big(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\big)\ dA$ where $C$ is a positively oriented, smooth, closed curve and let $D$ be the region bound by $C$ in the plane.
Therefore, my region $D$ is the ellipse $\frac{x^2}{(\sqrt{2})^2}+y^2=1$. My bounds for $D$ would then be $-\sqrt{2}\leq x \leq \sqrt{2},\space and \space 0\leq y \leq\sqrt{1-\frac{x^2}{2}}$.
I realize though that this is only the upper half of the ellipse, so I multiply my integral by $2$ $\therefore$
$$2\int_{-\sqrt{2}}^{\sqrt{2}}\int_0^{\sqrt{1-\frac{x^2}{2}}}-2y^3\ dy\ dx$$
However, after evaluating this integral, my result is $\frac{22\sqrt{2}}{15}$ when it should be $0$. Is it incorrect to multiply the integral by $2$ when I should instead be integrating in the $y$ direction from $$-{\sqrt{1-\frac{x^2}{2}}}\leq y \leq{\sqrt{1-\frac{x^2}{2}}}$$
Best Answer
You decided to use Green's theorem to compute the given line integral $J$, and arrived at $$J=\int_D-2y^3\>{\rm d}(x,y)\ ,\tag{1}$$ where $D:=\{(x,y)\,|\,x^2+2y^2\leq2\}$ is the elliptical disc bounded by $C$. Now the integral $(1)$ is $=0$ by symmetry since $D$ is symmetric with respect to the $x$-axis, and the integrand $-2y^3$ is an odd function of $y$.
You can replace $\int_{-b}^b f(y)\>dy\ $ by $\ 2\int_0^b f(y)\>dy$ only if $f$ is an even function of $y$.