The problem is the following: given the function $f(x,y) = 3x + 2y$ and the domain $D = \{x^2 + 4y^2 \le 4\}$, find the values of $f$ on the boundary.
Here is my approach:
- Rewrite $f$ in polar coordinates. $$ f(x,y) = f(r, \theta) = 3r\cos(\theta)+2r\sin(\theta) = r(3\cos(\theta)+2\sin(\theta)) $$
- Rewrite the boundary of $D$ in polar coordinates. $$ x^2 + 4y^2 = x^2 + y^2 + 3y^2 = r^2 + 3r^2\sin^2(\theta)=4\\ r^2(1 + 3\sin^2(\theta))=4\\ r=\frac{2}{\sqrt{1 + 3\sin^2(\theta)}}, \theta\in[0, 2\pi]\\ $$
- Create a new function by plugging in the expression above into $f$. $$ g(\theta) = \frac{2}{\sqrt{1 + 3\sin^2(\theta)}}(3\cos(\theta)+2\sin(\theta)), \theta\in[0, 2\pi]\\ $$
The correct answer however is $$ 2\sin(\theta) + 6\cos(\theta), \theta\in[0, 2\pi] $$ Could someone please help me identify my mistake?