Problem
I have curve defined in polar coordinates in terms of radius:
$$ r=\sqrt{\sin(\theta)} $$ And i would like to compute half of it's area so $r \in [0,\pi]$
Attempt to solve
A point in polar coordinate system is defined with $p(r,\theta)$ radius and angle (in radians in this case.) A probably good solution to this would be to use two integrals. $$ \int_{0}^{\pi} \int_{0}^{\sqrt{sin{\theta}}} drd\theta $$ $$ =\int_{0}^{\pi}\sqrt{\sin{\theta}}d\theta $$ using u substitution $u=\sin(\theta)$, then $\frac{du}{d\theta}=\cos(\theta) \rightarrow du=\cos(\theta)d\theta \rightarrow d\theta =\frac{du}{\cos(\theta)}$ this would also change the integration bounds to $[-1,1]$ $$=\frac{1}{\cos(\theta)} \int_{-1}^{1}\sqrt{u}du $$ $$=\frac{1}{\cos{\theta}}(\frac{2u^{2/3}}{3}+C)\big|_{-1}^{1} $$ Now i would recieve $$ =\frac{1}{\cos(\theta)}\frac{4}{3} $$ which clearly cannot be the area of this graph in polar coordinates. I think i lack the intuition when it comes to integration in polar coordinates. I made some plots with matlab to get more intuition about this subject.
Plot of $r=\sqrt{sin(\theta)}$ in polar.
Now when i want to to compute area i split the whole area into smaller segments or (sectors in this case) like this.
Now what riemann integral does it computes the area when these triangles $d\theta \rightarrow 0$ the gap between each of these segments approaches zero in terms of the angle $\theta$. It would probably look something like this:
The gaps are so small it's hard to tell where each line is.
One thing i noticed right away is that sector area is $A=\frac{r^2\theta}{2}$
So i would actually want to integrate something like this:
$$ \int_{0}^{\pi} \frac{(\sqrt{\sin(\theta)})^2\theta}{2}d\theta$$ $$ \frac{1}{2}\int_{0}^{\pi} \theta \sin(\theta) d\theta $$ Now when integrating in parts: $$ (-\frac{1}{2}\theta \cos(\theta))\big|_{0}^{\pi} + \frac{1}{2} \int_{0}^{pi}\cos(\theta) d\theta $$ $$ =\frac{\pi}{2} $$
Now area of half of this should be $\frac{\pi}{2}$ and the whole area should be $2\frac{\pi}{2}=\pi$ because of symmetry.
Now i would like to have some feedback if my answer seems correct or not ?. Also if the reasoning seems correct / maybe there is flaw please let me know. That would be highly appreciated.
Edit
According to comments i should use $$ \frac{1}{2}\int_{0}^{\pi}r^2 d\theta $$ where $r=\sqrt{\sin(\theta)}$ I get $$ \frac{1}{2}\int_{0}^{\pi} (\sqrt{\sin(\theta)})^2d\theta$$ Now i think i see the reason why this is true. The reason i failed last time on this was that i didn't think formula of $A=\frac{1}{2}r^2\theta$. I didn't think of what it means. When i look at it now i can see that of course $d\theta$ is what i want to use instead of $\theta$ since we want to only compute are of this small sectors. And each small sector has $\theta$ of $d\theta$
The expression can be simplified now:
$$ \frac{1}{2} \int_{0}^{\pi} \sin(\theta) d\theta $$
Now this integrates nicely. Our integrand is only $\sin(\theta)$ (no need for integration in parts)
$$ \frac{1}{2} (-\cos(\theta)+C)\bigg|_{0}^{\pi}=\frac{1}{2}(-\cos(\pi)-(-\cos(0))=\frac{1}{2}(1+1)=1 $$
area of this surface is $1$ (which is half of the whole area). The whole are should be $2$ because of the symmetry.