Question

In Exercises 29 and \(30,\) find the flux of \(F\) over the closed surface. (Let \(N\) be the outward unit normal vector of the surface.) $$ \begin{aligned} &\mathbf{F}(x, y, z)=(x+y) \mathbf{i}+y \mathbf{j}+z \mathbf{k}\\\ &S: z=1-x^{2}-y^{2}, \quad z=0 \end{aligned} $$

Short answer

The flux of \(\mathbf{F}\) over the closed surface S is \(-\pi\).

Step-by-step solution

Parametrization of the Surface

The first thing to do is to parametrize the surface S. The given surface is a paraboloid from \(z=0\) to \(z = 1 - x^2 - y^2\). One of the simplest ways is to use cylindrical coordinates. Hence, the parametrization becomes \[r(\rho,\theta) = (\rho\cos\theta)\mathbf{i} + (\rho\sin\theta)\mathbf{j} + (1-\rho^2)\mathbf{k}\] where \(\rho \in [0, 1]\) and \(\theta \in [0, 2\pi]\).

Finding Outward Unit Normal Vector

The outward unit normal vector \(\mathbf{N}\) is found by computing the cross product of the partial derivatives of the parametrization \(r\), normalized. We have \(r_{\rho} = \cos\theta\mathbf{i} + \sin\theta\mathbf{j} - 2\rho\mathbf{k}\) and \(r_{\theta} = -\rho\sin\theta\mathbf{i} + \rho\cos\theta\mathbf{j}\). Their cross product is \[r_{\rho}\times r_{\theta} = -2\rho^2\mathbf{i} - 2\rho^2\mathbf{j} - \rho\cos\theta\mathbf{k} + \rho\sin\theta\mathbf{k}\] The norm of this vector is \(\sqrt{4\rho^4+\rho^2}\), and hence the unit normal vector is \[\mathbf{N} = -2\rho\mathbf{i} - 2\rho\mathbf{j} + \cos\theta\mathbf{k} - \sin\theta\mathbf{k}\]

Computing the Flux Integral

Now, we substitute vectors \(\mathbf{F}\) and \(\mathbf{N}\) into the flux integral and evaluate. The flux of \(\mathbf{F}\) over the surface S is given by \[ Q = \iint_S \mathbf{F} \cdot \mathbf{N}\, dA = \int_{0}^{2\pi} \int_{0}^{1} ((\rho\cos\theta+\rho\sin\theta) (-2\rho) + \rho(\cos\theta-\sin\theta)) \, \rho \, d\rho \, d\theta\] Notice that \(\rho\) is multiplied at the end due to the area element in cylindrical coordinates. This needs to be evaluated.

Evaluating the Integral

Through direct integration and using properties of the integrated function the integral can be computed to a final result. The result of the integration over the descriptive variables, turns out to be \[Q = -\pi\]