Triple integration in cylindrical coordinates Determine the value of $ \int_{0}^{2} \int_{0}^{\sqrt{2x - x^2}} \int_{0}^{1} z \sqrt{x^2 +y^2} dz\,dy\,dx $
My attempt: So in cylindrical coordinates, the integrand is simply $ \rho$.
$\sqrt{2x-x^2} $ is a circle of centre (1,0) in the xy plane. So $ x^2 + y^2 = 2x => \rho^2 = 2\rho\cos\theta => \rho = 2\cos\theta $
Therfore, I arrived at the limit transformations, $ 0 < \rho < 2\cos\theta,\,\, 0 < z < 1, \text{and}\,\,0 < \theta < \frac{\pi}{2}  $
Bringing this together gives $  \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta} \int_{0}^{1} z\,\,\rho^3\,dz\,d\rho\,d\theta $ in cylindrical coordinates. Is this correct?

As joriki noted completely; your integral would be $$  \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta} \int_{0}^{1} z\,\,\rho^2\,dz\,d\rho\,d\theta=\bigg(\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta}\rho^2\,d\rho\,d\theta \bigg)\times \int_{0}^{1} z\ dz\\\ =\bigg(\int_{0}^{\frac{\pi}{2}} \frac{\rho^3}{3}\bigg|_0^{2\cos(\theta)}d\theta \bigg)\times \frac{1}{2}=\frac{8}{9}$$
