How do I calculate cross-sectional area of this:
${r}=\sqrt{sin\theta} \, $, when $ 0 \le \theta \le \pi,$
Don't know what is the right answer but I have get that the area is 1. Is that right answer?
$\int_0^\pi\int_0^\sqrt{sin\theta} r\, \textrm{d}r\textrm{d}\theta.$ Is this right?
This is what I did and the answer I got: $$\int_\limits{0}^\pi[\frac{r^2}2]_0^{\sqrt{\sin(\theta)}}]d\theta$$
$$\int_\limits{0}^\pi(\frac{\sin\theta}2)d\theta$$
$$[\frac{-\cos{\theta}}{2}]_0^\pi$$
$$-\frac{\cos\pi}{2}- -\frac{\cos(0)}{2}=.5+.5=1$$