Let $C$ be the curve parameterized by $r(t)=(5+2 \cos t, 2 \sin t)$ for $0≤t≤2π$. I'm trying to evaluate $\oint_{C} (\cos x e^x + y)dx$ over this region.
I tried to use Green's Theorem to solve this:
$$\int (\cos x e^x + y)dx$$
$$=\iint\bigg(\frac{\partial(0)}{\partial x} -\frac{\partial(\cos (x) e^x + y)}{\partial y}\bigg)dxdy$$
$$=\iint(0-1)dxdy$$
$$=- \iint dxdy$$
However, I'm getting stuck on the bounds of this integral now and how to express the curve $C$. I know the curve is an ellipse, but I'm having trouble representing it in either Cartesian or polar coordinates. Any guidance is greatly appreciated!
$$r(t)=(5+2 \cos t, 2 \sin t)$$
Well here is the deal, since we are dealing in cartesian components $r(t) = <x,y>$ , by equating vector components we have:
$$ x = 5 + 2 \cos t$$ $$ y = 2 \sin t$$
Now, to recover the region, we need the cartesian curve of the domain, we can manipulate the above pairs of equation to remove the $t$ parameter dependence. This gives a circle.
Change of variables?
$$ x-5 = r \cos \theta$$
$$ y= r \sin \theta$$
We can use the above relation on the original set of two equations in cartesian with paramter to get:
$$ 2 \cos t = r \cos \theta \tag{1}$$
$$ 2 \sin t =r \sin \theta \tag{2}$$
Square and add the two equations above:
$$ 4 = r^2$$
By division we see $\tan t = \tan \theta$ , we can find that $\theta$ can take anything from $ \left[ 0 , 2 \pi \right]$. Now, we just need to find the area element, this we can do by Jacobian/ Differential forms.
$$ dx dy = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \\ \end{vmatrix} dr d \theta= r^2 dr d \theta$$ from (1) and (2).
Our integral now is:
$$ \int \int r^2 dr d \theta$$
And our domain is the region enclosed in $r^2 = 4 $, which is $r<2$ and $ 0 < \theta < 2 \pi$. Since each of these variables are seperable, we have the integral:
$$ \int_0^{2 \pi} d \theta \int_0^2 r^2 dr$$