91Ó°ÊÓ

Recall that the gradient of a scalar function is given by \(\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle\). We have \(\ln |\mathbf{r}| = \ln \sqrt{x^2 + y^2 + z^2}\). Let's compute the gradient: \(\nabla \ln |\mathbf{r}| = \langle \frac{\partial}{\partial x} \ln \sqrt{x^2 + y^2 + z^2}, \frac{\partial}{\partial y} \ln \sqrt{x^2 + y^2 + z^2}, \frac{\partial}{\partial z} \ln \sqrt{x^2 + y^2 + z^2} \rangle\) After simplification, we get: \(\nabla \ln |\mathbf{r}| = \langle \frac{x}{x^2 + y^2 + z^2}, \frac{y}{x^2 + y^2 + z^2}, \frac{z}{x^2 + y^2 + z^2} \rangle\)

The hemisphere \(S\) can be parametrized by the following function: \(G(\theta, \phi) = \langle a\sin(\phi) \cos(\theta) , a\sin(\phi) \sin(\theta), a\cos(\phi) \rangle\) Hence, \(x = a\sin(\phi) \cos(\theta)\), \(y = a\sin(\phi) \sin(\theta)\), and \(z = a\cos(\phi)\). The domain for integration will be \(\theta \in [0, 2\pi]\), and \(\phi \in [0, \frac{\pi}{2}]\). Compute the tangent vectors for the parametrization: \(\frac{\partial G}{\partial \theta} = \langle -a\sin(\phi)\sin(\theta), a\sin(\phi)\cos(\theta), 0 \rangle\) \(\frac{\partial G}{\partial \phi} = \langle a\cos(\phi)\cos(\theta), a\cos(\phi)\sin(\theta), -a\sin(\phi) \rangle\) Now we can compute the normal vector by taking cross product of these tangent vectors: \(\mathbf{n} = \frac{\partial G}{\partial \theta} \times \frac{\partial G}{\partial \phi}\) \(\mathbf{n} = \langle a^2 \sin^2(\phi) \cos(\theta), a^2 \sin^2(\phi) \sin(\theta), a^2 \sin(\phi) \cos(\phi) \rangle\)

Substitute \(x\), \(y\), and \(z\) from parametrization into the gradient: \(\nabla \ln |\mathbf{r}| = \langle \frac{\sin(\phi) \cos(\theta)}{a}, \frac{\sin(\phi) \sin(\theta)}{a}, \frac{\cos(\phi)}{a} \rangle\) Now, we can take the dot product of gradient and normal vector and set the integral to find the answer: \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (\nabla \ln |\mathbf{r}| \cdot \mathbf{n}) a^2 \sin(\phi) d \phi d \theta\) After taking the dot product and simplifying, we get: \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = a^2 \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (1 + \sin^2(\phi)) \sin(\phi) d \phi d \theta\) Now, integrate with respect to \(\phi\) and \(\theta\): \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = a^2 [|[-\cos(\phi) + \frac{1}{3}\cos^3(\phi)]_{0}^{\frac{\pi}{2}}| \int_{0}^{2\pi} d \theta\) This simplifies to: \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = a^2 (\frac{4}{3}) [2\pi - 0]\) Finally, the integral evaluates to: \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = \frac{8}{3} a^2 \pi\)