Surface Integrals of Vector Fields — Question 3

Question 3 – Solution

Strategy. Use angle and height parameters. The outward cross product is horizontal, so the vertical component of the field contributes nothing.

Step 1: Parametrize $$\mathbf r(\theta,z)=\langle 2\cos\theta,2\sin\theta,z\rangle, \quad 0\le\theta\le 2\pi,\quad 0\le z\le 3.$$ The ordered cross product $$\boxed{\mathbf r_\theta\times\mathbf r_z =\langle 2\cos\theta,2\sin\theta,0\rangle}$$ points away from the axis.

Step 2: Form the integrand On $S$, $$\mathbf F=\langle 2\cos\theta,2\sin\theta,z^2\rangle,$$ and therefore $$\mathbf F\cdot(\mathbf r_\theta\times\mathbf r_z) =4\cos^2\theta+4\sin^2\theta=4.$$

Step 3: Integrate $$\boxed{\iint_S\mathbf F\cdot\mathbf n\,dS =\int_0^{2\pi}\int_0^3 4\,dz\,d\theta=24\pi}.$$

Verification The flux density is constantly positive, so the field points outward through every point of the curved side. The $z^2$ component is tangent to that side and correctly disappears.