I need to calculate the surface of the area between $$\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} \le 1$$ and $$y \ge 0$$
Let's introduce polar coordinates:
$$x-2 = 3r\cos(\phi)$$ $$y+1 = 2r\sin(\phi)$$ $$|J| = 6r$$
Substituting this in the first inequality we get that $r^2 \le 1$, which means $r \in [0,1]$ for now.
Now here's the problem. In the second inequality we have:
$$r\sin(\phi) \ge 0$$
As $r$ is already greater or equal to zero, we have that $\sin(\phi) \ge 0$, and thus that $\phi \in [0, \pi]$
One can also see this by drawing a picture, the line $y=0$ is the x axis and we get that it's greater than zero above the x axis, which is from $0$ to $\pi$
Substituting this, I get the following double integral:
$$ \int_{0}^{\pi}\int_{0}^{1} 6rdrd\phi$$ and when I solve it I get that the solution is $3\pi$. However, my workbook states that the solution is $2\pi - \frac{3\sqrt{3}}{2}$
Can anyone help me understand where I might be wrong?
Please note that
$y = 2 r \sin \phi - 1 \geq 0 \implies r \geq \dfrac{1}{2\sin\phi}$. So the lower bound of $r$ is not zero as you are between a line and the circumference.
Also to find bounds of $\phi$, plug in $r = 1$.
So, $\sin \phi = \dfrac{1}{2} \implies \dfrac{\pi}{6} \leq \phi \leq \dfrac{5\pi}{6}$