Let \[ S=\{(x,y,z)\in\RR^3:x^2+y^2+z^2=1\}. \] Evaluate \[ \iint_S (x^2+y+z)\,dA. \]
Solution
1. The linear terms have zero surface integral:\[ \iint_S y\,dA=0, \qquad \iint_S z\,dA=0. \]
Reflection across the plane \(y=0\) preserves the sphere and its area measure while replacing \(y\) by \(-y\). Hence the integral of \(y\) is its own negative and is therefore zero. Reflection across \(z=0\) gives the same argument for \(z\).
2. One has\[ \iint_S x^2\,dA=\frac{4\pi}{3}. \]
Rotations permuting the coordinate axes preserve \(S\) and its area measure, so \[ \iint_S x^2\,dA = \iint_S y^2\,dA = \iint_S z^2\,dA. \] Call this common value \(I\). Since \[ x^2+y^2+z^2=1 \] on \(S\), \[ 3I = \iint_S(x^2+y^2+z^2)\,dA = \iint_S1\,dA = \operatorname{area}(S) = 4\pi. \] Thus \(I=4\pi/3\).
3. Therefore\[ \boxed{ \iint_S(x^2+y+z)\,dA=\frac{4\pi}{3}. } \]
Combine steps 1 and 2.
4. Q.E.D.
Step 3 is the requested value.