Surface Integrals of Vector Fields — Question 4

Question 4 – Solution

Strategy. On a centered sphere, the field and unit normal are parallel; only their magnitudes need to be compared.

Step 1: Normal component Write $\mathbf x=\langle x,y,z\rangle$. On $S_a$, $$\mathbf n=\frac{\mathbf x}{a},\qquad \mathbf F=\frac{\mathbf x}{a^3}.$$ Consequently, $$\boxed{\mathbf F\cdot\mathbf n =\frac{\mathbf x\cdot\mathbf x}{a^4}=\frac 1{a^2}}.$$

Step 2: Integrate Since the sphere has area $4\pi a^2$, $$\boxed{\iint_{S_a}\mathbf F\cdot\mathbf n\,dS =\frac 1{a^2}(4\pi a^2)=4\pi}.$$

Step 3: Interpret the scale Increasing the radius weakens the normal component by the factor $a^{-2}$ while increasing spherical area by $a^2$. These factors cancel exactly.

Verification The field is everywhere outward on $S_a$, so the positive sign is correct. The calculation is direct and uses no theorem about closed surfaces.