Set $D$ is bounded by the lines $y=0,x=0$ and by the curve (in polar coordinates) $r(\theta)=e^{\theta}, 0 \le \theta \le \frac{\pi}{2}$.
We define $Q=\{(x,y,z)\in\mathbb{R}^3: (x,y)\in\partial D, 0 \le z \le \sqrt{x^2+y^2} \}$. Find the area of $Q$.
Note: $\partial D$ is boundary of $D$.
I'm having a difficult time deciding what to do in this question, having the curve in polar coordinates is very confusing.
If I want to move to polar coordinates: $x=pcos(t), y=psin(t)$, then what about the curve $r(\theta)$?
I'm really stuck here and I would appreciate any help.
Thanks in advance!
Edit (Attempt after help of Math Lover):
$x=e^{\theta}\cos(\theta), y=e^{\theta}\sin(\theta), z=z$.
Finding the bounds:
$0\le z\le\sqrt{e^{2\theta}cos^2\theta+e^{2\theta}sin^2\theta}=e^{\theta}$.
$e^{\theta}cos\theta=0 \Longrightarrow \theta = \frac{\pi}{2}$
$e^{\theta}sin\theta=0 \Longrightarrow \theta=0.$
(Was this necessary? I got the same bounds for $\theta$).
Now I know that in order to find area of a surface, I need to have the "upper" curve, which is $z=e^{\theta}=f(x(\theta),y(\theta))$.
So, a vector that draws my curve:
$\vec l=(e^{\theta}cos\theta, e^{\theta}sin\theta, z), 0 \le z \le e^{\theta}. 0 \le \theta \le \frac{\pi}{2}$.
But here I got stuck, otherwise I'll get a double integral, I'm thinking of why can I remove $z$ or why not.
Completing the solution:
$\vec l = (e^{\theta}cos\theta, e^{\theta}sin\theta)$
$d\vec l = (-e^{\theta}sin\theta + e^{\theta}cos\theta, e^{\theta}cos\theta + e^{\theta}sin\theta)$.
$|d\vec l| = \sqrt{(-e^{\theta}sin\theta + e^{\theta}cos\theta)^2+( e^{\theta}cos\theta + e^{\theta}sin\theta))^2}=\sqrt{e^{2\theta}-e^{2\theta}sin(2\theta) + e^{2\theta}+e^{2\theta}sin(2\theta)}=\sqrt{2}e^{\theta}$ (Yay!)
$$\int_{l}fdl=\int_{0}^{\frac{\pi}{2}}e^{\theta}\sqrt{2}e^{\theta}d\theta = \sqrt{2}[\frac{e^{\pi}}{2}-\frac{e^0}{2}]$$