My task is as follows:
$$\: \mathcal{C}:\textbf{r}(t) = \big(1 + 2\cos(t)\big)\:\hat{i} + \big(3\sin(t) - 2\big)\: \hat{j},\enspace t\in [0, 2\pi].$$
Calculate $\oint_\mathcal{C} \textbf{F}\:d\textbf{r}$ when: $$\textbf{F}(x,y) = y^2\:\hat{i} + x\: \hat{j}.$$
My work goes as follows;
The definition is: $$\oint_\mathcal{C} \textbf{F}\:d\textbf{r} = \int_a^b \textbf{F}\left(\textbf{r}(t)\right)\textbf{r}'(t)\:dt = \iint_R \left( Q_x - P_y\right)\:dxdy.$$
Where $Q_x, P_y$ are partial derivatives.
With this at hand we have: $$\textbf{r}'(t) = -2\sin(t)\: \hat{i} \:+ 3\cos(t)\:\hat{j},\enspace \textbf{F}\left(\textbf{r}(t)\right) = \left(3\sin(t) - 2\right)^2\:\hat{i} + \left(1 + 2\cos(t) \right)\:\hat{j}.$$
And: $$\textbf{F}\left(\textbf{r}(t)\right)\textbf{r}'(t) = 3\cos(t) + 6\cos^2(t) - 18\sin^3(t) + 24\sin^2(t) - 8\sin(t).$$
While i think i've calculated this right, i get the feeling that this could be done in a much easier way. I would like to try the last step of the definition, but im, not sure how to set up those boundaries with a non-trivial ellipse like this i.e. where center $S(1,-2)$ and radiuses $a = 2, b = 3.$
Any help would be appreciated!
Thanks in advace.
Yes, this is more manageable doing the integral over the ellipse.
You can use some changes of variables to integrate over the ellipse $(\frac{x-1}{2})^2+(\frac{y+2}{3})^2\le 1$. First take $u=\frac{x-1}{2}$ $v=\frac{y+2}{3}$.
Then you have:
$\int\int_R (Q_x-P_y)dxdy=\int\int_R (1-2y)dxdy=\int\int_D (5-6v)6dudv$
where $D$ is the unit disc.
Then use polar coordinates $u=r\cos\theta, v=r\sin\theta$ so you get:
$6\int_0^1\int_0^{2\pi} (5-6r\sin\theta)rd\theta dr=30\pi$