Triple integration in spherical polars

[CAF123]

Homework Statement

Determine the value of $$\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \int_{0}^{\sqrt{1-x^2-y^2}} \sqrt{x^2+ y^2 + z^2} dz dy dx$$

The Attempt at a Solution

So in spherical polars, the integrand is simply ρ.
$\sqrt{1- x^2- y^2} = z = ρ\cos\phi = \cos\phi$ since we are on the unit sphere.
This gives one of the bounds
$\sqrt{1-x^2}$ is the upper half of the unit circle in the xy plane, so clearly θ goes from 0 to pi.
Since we consider z≥0, $\phi$ must go from 0 to pi/2.

Putting this together gives, $$\int_{0}^{\frac{π}{2}} \int_{0}^{π} \int_{0}^{cos\phi} ρ\,dρ\,dθ\,d\phi$$ have I ordered the integration process correctly?

 

[Fightfish]

I'm afraid it's totally wrong.

It can be tricky changing the limits of integration when changing from one set of coordinates to another. It would be easier to visualise the region R and then parameterise it in terms of the new coordinate system.

The region of integration is the volume of the unit sphere centred at the origin in the first quadrant (ie x, y and z are all positive). This region can be parameterised in spherical coordinates as:
$$R: \{0\leq \rho \leq 1, 0\leq\theta\leq\frac{\pi}{2}, 0\leq\phi\leq\frac{\pi}{2}\}$$
Furthermore, note that $dx\,dy\,dz \neq d\rho\,d\theta\,d\phi$
The infinitesimal volume element in spherical coordinates is in fact
$$dV = \rho^{2} sin \phi \,d\rho\,d\theta\,d\phi$$
($\rho^{2} sin \phi$ is often called the Jacobian)

 

[CAF123]

Thanks for clarifying things.
It is clear that ρ goes from 0 to 1 when I see the picture, but why don't I get the same results when I tried to do it algebraically? (I get $\cos\phi$)
Also , should I expect the answer to be pi/6, ie a quarter of the volume of a unit hemisphere?