# Triple integration problem by changing to polar co-ordinate

I have this question

Evaluate the triple integral: $$\int_0^1 \int_0^{\sqrt (1-x^2)} \int_0^{\sqrt (1-x^2-y^2)} xyz~dx~dy~dx$$

My attemp: Changing to polar co-ordinates.
Let \begin{align} x=rsin\theta cos\phi \\ y=rsin\theta sin \phi \\ z=rcos\theta \end{align} The new integrals becomes $$\int_0^1 \int_{\theta=0}^{\pi/2} \int_{\phi=0}^{\pi/2} rsin\theta cos\phi rsin\theta sin \phi rcos\theta|J|~dx~dy~dx$$ where $|J|$ is the Jacobian of transformation. Evaluating this, gives me the value of the integral as $\frac 1{144}$. However the answer is $\frac 1{48}$. Where did I go wrong?

The Jacobian is $r^2 \sin{\theta}$, so the integral is
$$\underbrace{\int_0^1 dr \, r^5}_{1/6} \underbrace{\, \int_0^{\pi/2} d\theta \, \sin^3{\theta} \cos{\theta}}_{1/4} \, \underbrace{\int_0^{\pi/2} d\phi \, \cos{\phi} \sin{\phi}}_{1/2}$$