Evaluate the surface integral ∫∫ (3x i + 2y j). dS, where S is the sphere given by x2 + y2 + z2 = 9

To evaluate the surface integral (iint (3xmathbf{i} + 2ymathbf{j}) cdot dmathbf{S}), where (S) is the sphere given by (x^2 + y^2 + z^2 = 9), we use the fact that the sphere has radius (r=3) and is centered at the origin.

Given the vector field (mathbf{F} = 3xmathbf{i} + 2ymathbf{j}), notice that the vector field’s third component is zero (F_z=0), implying that it has no component in the (z)-direction.

The surface integral over a closed surface, like a sphere, can be computed via the divergence theorem. However, in this specific case, calculating the vector field’s dot product with the outward normal directly and integrating over the surface might not be straightforward due to the absence of the (z)-component in (mathbf{F}). Nonetheless, it is more insightful to exploit the symmetry of the sphere and the nature of the vector field for this calculation.

Symmetry Insight:
1. For the component (3xmathbf{i}), its effect cancels out symmetrically in the integral over the sphere because for every (x), there is a (-x) with equal contribution but opposite directions when projected to the surface area element (dmath