# 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?

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Limits look fine, but $\rho^3$ should be $\rho^2$. – Hans Lundmark Oct 21 '12 at 11:24

It all seems correct except it should only be $\rho^2$, since you had one factor of $\rho$ in the integrand and the Jacobian for cylindrical and polar coordinates only contains one factor of $\rho$; it's the one for spherical coordinates that has two. (You can remember this by noting that the factors of $\rho$ need to compensate the number of coordinates that get replaced by angles for the units to come out right.)
@CAF: You can change the order of integration of any variables whose limits don't depend on each other, so yes, you can integrate over $\rho$, then $z$ and then $\theta$, but the order isn't fully arbitrary; you can't integrate over $\theta$ first because the limit for $r$ depends on $\theta$. No, the answer isn't $1/3$; Babak carried out the integration in his answer. – joriki Oct 21 '12 at 17:34
As joriki noted completely; your integral would be $$\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta} \int_{0}^{1} z\,\,\rho^2\,dz\,d\rho\,d\theta=\bigg(\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta}\rho^2\,d\rho\,d\theta \bigg)\times \int_{0}^{1} z\ dz\\\ =\bigg(\int_{0}^{\frac{\pi}{2}} \frac{\rho^3}{3}\bigg|_0^{2\cos(\theta)}d\theta \bigg)\times \frac{1}{2}=\frac{8}{9}$$