<s>[INST] Area between two trigonometric curves I need to find the area between two curves:
$$\begin{cases} x=\sqrt { 2 } \cos { t }  \\ y=4\sqrt { 2 } \sin { t }  \end{cases}\\ y=4\quad (y\ge 4)$$
I came up with: 
$$\frac { 1 }{ 4 } \left( 8\pi n +\pi  \right) \le t\le \frac { 1 }{ 4 } \left( 8\pi n + 3\pi  \right) $$
$$\int _{ \frac { 1 }{ 4 } (8\pi n+\pi ) }^{ \frac { 1 }{ 4 } (8\pi n+3\pi ) }{ 4\sqrt { 2 } \sin { (t) }  } -\sqrt { 2 } \cos { (t) } \quad dx =2\left( \sin { (2\pi n) } +4\cos { (2\pi n) }  \right)$$
So the area is a function of $n$, though I was supposed to get a finite solution. What am I doing wrong?
 [/INST] As I said in my comment, you need to restrict your angles, both in general, to have only one revolution around the origin, and when you want to meet your additional restriction on $y$. For the first case, it's obvious that $t \in [0, 2\pi)$. For the second one, solve an easy equation
$$
y = 4 \implies 4\sqrt 2 \sin t = 4 \implies \sin t = \frac 1{\sqrt 2} \implies t_1 = \frac \pi 4,\ t_2 = \frac {3\pi}4
$$
If you visualize the analysis above, you get

You need to find the area of the shape with red dome and black straight base, 
Finding the area of the red curve and $y = 0$ is as easy as 
$$
A_f = \int_{\frac {3\pi} 4}^{\frac \pi 4} y(t)\ x'(t)\ dt = 8 \int_{\frac \pi 4}^{\frac {3\pi}4} \sin^2 t\ dt = 4 \left . \left( t - \frac {\cos 2t}2\right) \right |_{\frac \pi 4}^{\frac {3\pi}4} = 2(2 + \pi)
$$
since you know your angles.
And the area you need is the difference between the area above and rectangle with dashed sides, solid black top and bottom side on $x$ axis, which is
$$
A_r = 2 \cdot 4 = 8
$$
and finally, $A = A_f - A_r = 2\pi - 4$
PS
In integration, I used the positive direction of $x$ to put upper and lower bounds for $t$, but then because of the negative sign that comes from $x'(t)$ I switched them again to get increasing order of angles.
</s>