To evaluate the integral \(\int_0^1 \int (x^2 + y^2 + z^2) \, dxdydz \) over the volume enclosed by the surface \( x^2 + y^2 + z^2 = 1 \), we can use spherical coordinates to simplify the computation. Here is a detailed, step-by-step solution:
1. Express the integrand and limits in spherical coordinates:
[tex]\[
x = r \sin \theta \cos \phi, \quad y = r \sin \theta \sin \phi, \quad z = r \cos \theta
\][/tex]
The volume element in spherical coordinates is given by:
[tex]\[
dV = r^2 \sin \theta \, dr \, d\theta \, d\phi
\][/tex]
2. Rewrite the integrand \(x^2 + y^2 + z^2\) in spherical coordinates:
[tex]\[
x^2 + y^2 + z^2 = (r \sin \theta \cos \phi)^2 + (r \sin \theta \sin \phi)^2 + (r \cos \theta)^2
\][/tex]
Simplifying this, we get:
[tex]\[
x^2 + y^2 + z^2 = r^2 (\sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi + \cos^2 \theta) = r^2 (\sin^2 \theta (\cos^2 \phi + \sin^2 \phi) + \cos^2 \theta) = r^2 (\sin^2 \theta + \cos^2 \theta) = r^2
\][/tex]
3. Set up the integral in spherical coordinates:
[tex]\[
\iiint_V (x^2 + y^2 + z^2) \, dV \rightarrow \iiint_V r^2 \cdot r^2 \sin \theta \, dr \, d\theta \, d\phi = \iiint_V r^4 \sin \theta \, dr \, d\theta \, d\phi
\][/tex]
The limits for the spherical coordinates are:
[tex]\[
0 \leq r \leq 1, \quad 0 \leq \theta \leq \pi, \quad 0 \leq \phi \leq 2\pi
\][/tex]
4. Evaluate the integral step-by-step:
First, integrate with respect to \(\phi\):
[tex]\[
\int_0^{2\pi} \int_0^{\pi} \int_0^1 r^4 \sin \theta \, dr \, d\theta \, d\phi
\][/tex]
The \(\phi\) part is straightforward:
[tex]\[
\int_0^{2\pi} d\phi = 2\pi
\][/tex]
Now, perform the integration with respect to \(r\):
[tex]\[
\int_0^1 r^4 \, dr = \left[ \frac{r^5}{5} \right]_0^1 = \frac{1}{5}
\][/tex]
Finally, perform the integration with respect to \(\theta\):
[tex]\[
\int_0^\pi \sin \theta \, d\theta = \left[ -\cos \theta \right]_0^\pi = (-\cos \pi) - (-\cos 0) = 1 - (-1) = 2
\][/tex]
5. Combine the results:
[tex]\[
(2\pi) \cdot \left( \frac{1}{5} \right) \cdot (2) = 2\pi \cdot \frac{2}{5} = \frac{4\pi}{5}
\][/tex]
Therefore, the final result is:
[tex]\[
\boxed{1.67551608191456}
\][/tex]
