Question

In Exercises \(23-28,\) find the flux of \(F\) through \(S\), \(\int_{S} \int \mathbf{F} \cdot \mathbf{N} d S\) where \(\mathbf{N}\) is the upward unit normal vector to \(S\). $$ \mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}-2 z \mathbf{k} ; S: z=\sqrt{a^{2}-x^{2}-y^{2}} $$

Short answer

The flux of the vector field F through the surface S is \(2/3 \pi a^{4}\).

Step-by-step solution

Parameterize the Surface S

Begin by parameterizing \(S\). Given \(S\) is defined as \(z=\sqrt{a^{2}-x^{2}-y^{2}}\), this is a hemisphere centered at the origin. Convert to spherical coordinates: \(r = a, \theta \in [0, 2 \pi], \phi \in [0, \frac{\pi}{2}]\) because the hemisphere is above the xy-plane. The parameterized surface is then \(\mathbf{r}(\theta, \phi) = a\sin\phi\cos\theta \mathbf{i} + a\sin\phi\sin\theta \mathbf{j} + a\cos\phi \mathbf{k}\).

Compute the Unit Normal Vector and the Differential dS

The unit normal vector to the surface is given by \(\mathbf{N} =\frac{\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}}{|\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}|}\). First, differentiate \(\mathbf{r}(\theta, \phi)\) with respect to \(\theta\) and \(\phi\) and cross them. Then, compute the magnitude of the cross product which will be \(|\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}| = a^{2} \sin\phi\). Thus, the unit normal vector will be \(\mathbf{N} = \sin\phi\cos\theta \mathbf{i} + \sin\phi\sin\theta \mathbf{j} + \cos\phi \mathbf{k}\). Thereafter, find the differential of the surface element dS: \(dS = |\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}| d\theta d\phi = a^{2}\sin\phi d\theta d\phi\).

Compute the Dot Product of Vector Field F and Unit Normal Vector N

Now, calculate the dot product of \(\mathbf{F}\) and \(\mathbf{N}\). Rewrite the vector field in spherical coordinates: \(\mathbf{F} = r\sin\phi\cos\theta \mathbf{i}+r\sin\phi\sin\theta \mathbf{j}-2r\cos\phi \mathbf{k}\). Substitute \(r = a\) because every point on the surface of the hemisphere is at distance \(a\) from the origin. The dot product is then \(\mathbf{F} \cdot \mathbf{N} = a^{2}\sin^{2}\phi(\cos^{2}\theta+\sin^{2}\theta)-2a^{2}\cos\phi\sin\phi\).

Evaluate the Surface Integral

Finally, substitute \(\mathbf{F} \cdot \mathbf{N}\) and \(dS\) into the surface integral and evaluate the integral. \(\int_{S} \int \mathbf{F} \cdot \mathbf{N} dS = \int_{0}^{2\pi}\int_{0}^{\pi/2} (a^{2}\sin^{2}\phi-2a^{2}\cos\phi\sin\phi) a^{2}\sin\phi d\phi d\theta = 2\pi a^{4}\int_{0}^{\pi/2} \sin^{3}\phi - 2\sin^{2}\phi \cos\phi d\phi\). Solve the above integral using trigonometric identities and integration techniques.