Surface Integrals of Vector Fields — Question 7

Question 7 – Solution

Strategy. A constant vertical field dotted with the upward graph element leaves exactly the projected area element.

Step 1: Graph orientation Write $$z=g(x,y)=\sqrt{4-x^2-y^2}$$ over the disk $D:x^2+y^2\le 4$. The outward orientation on the upper hemisphere is upward, so $$\mathbf n\,dS=\langle-g_x,-g_y,1\rangle\,dA.$$

Step 2: Project Since $\mathbf F=\mathbf k$, $$\mathbf F\cdot\mathbf n\,dS =\mathbf k\cdot\langle-g_x,-g_y,1\rangle\,dA=dA.$$ Thus $$\boxed{\iint_S\mathbf F\cdot\mathbf n\,dS =\iint_DdA=\pi(2^2)=4\pi}.$$

Step 3: Reverse orientation With the inward orientation the vector element changes sign, so the flux is $\boxed{-4\pi}$.

Verification The graph derivatives become unbounded at the equator, but their components are annihilated by the dot product with $\mathbf k$. Truncating to radius $R<2$ and taking $R\to 2^-$ yields the same result.