Use spherical coordinates to evaluate the triple integral of x^2 + y^2 + z^2 over the unit ball x^2 + y^2 + z^2 <= 1.

Step-by-step solution

1. 1

   Notice that both the integrand and the region are radially symmetric. The integrand $x^2+y^2+z^2$ and the boundary $x^2+y^2+z^2=1$ depend on nothing but distance from the origin. That is the definitive signal to use spherical coordinates: in Cartesian form this integral needs nested square-root limits, while in spherical form the limits are three constants.

2. 2

   Set up the spherical substitution, including the Jacobian. With

   $x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi$

   the integrand collapses to $x^2+y^2+z^2=\rho^2$, and the volume element carries the Jacobian factor

   $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$

   The $\rho^2\sin\phi$ is not optional bookkeeping — it is the local volume-stretching factor of the coordinate change, and omitting it is the most common way this integral goes wrong.

3. 3

   Write the limits. The unit ball is described by

   $0\le\rho\le1,\qquad 0\le\phi\le\pi,\qquad 0\le\theta\le2\pi$

   Note $\phi$ (the polar angle, measured down from the $+z$ axis) runs only to $\pi$, while $\theta$ (the azimuthal angle) runs the full $2\pi$. Swapping those ranges would double-count the ball.

4. 4

   Assemble the integral and separate it. Combining the integrand with the Jacobian gives $\rho^2\cdot\rho^2\sin\phi=\rho^4\sin\phi$, so

   $\iiint_B\left(x^2+y^2+z^2\right)dV=\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{1}\rho^4\sin\phi\,d\rho\,d\phi\,d\theta$

   Because the integrand factors as (function of $\rho$) $\times$ (function of $\phi$) $\times$ (constant in $\theta$) and the limits are constants, the triple integral splits into a product of three single integrals.

5. 5

   Evaluate the three factors.

   $\int_0^{2\pi}d\theta=2\pi,\qquad \int_0^{\pi}\sin\phi\,d\phi=\big[-\cos\phi\big]_0^{\pi}=1+1=2,\qquad \int_0^{1}\rho^4\,d\rho=\left[\frac{\rho^5}{5}\right]_0^1=\frac15$

   Multiplying:

   $(2\pi)(2)\left(\frac15\right)=\frac{4\pi}{5}\approx2.5133$

6. 6

   Sanity-check the size of the answer. The ball has volume $\tfrac43\pi\approx4.19$ and the integrand $\rho^2$ ranges from $0$ at the centre to $1$ at the surface, so the integral must lie strictly between $0$ and $4.19$. The mean value of $\rho^2$ over the ball is $\tfrac{4\pi/5}{4\pi/3}=\tfrac35$, which is sensibly above $\tfrac12$ because most of a ball's volume sits near its outer shell. A Monte Carlo estimate over the enclosing cube returns $\approx2.515$, confirming $4\pi/5$.