Consider a sphere of radius \(r=4\) centered at \((0,0,3)\). Let \(S_{1}\) be that portion of the spherical surface that lies above the \(x y\) plane. Find \(\int_{S_{1}}(\nabla \times \mathbf{H}) \cdot d \mathbf{S}\) if \(\mathbf{H}=3 \rho \mathbf{a}_{\phi}\) in cylindrical coordinates.

First, we need to convert the given vector field \(\mathbf{H} = 3\rho \mathbf{a}_{\phi}\) from cylindrical coordinates to Cartesian coordinates. The conversion formulas for cylindrical coordinates are as follows: \(x = \rho \cos \phi\) \(y = \rho \sin \phi\) In the cylindrical coordinate system, \(\mathbf{a}_{\phi}\) represents the unit vector in the \(\phi\) direction. The corresponding Cartesian representation of this unit vector is: \(\mathbf{a}_{\phi} = -\sin \phi \mathbf{i} + \cos \phi \mathbf{j}\) Now let's convert \(\mathbf{H}\): \[\mathbf{H}=3\rho(-\sin \phi\mathbf{i}+\cos \phi\mathbf{j})\] To write \(\mathbf{H}\) in terms of \(x\) and \(y\), we use the relationships between \(x\), \(y\), and \(\rho\), \(\phi\). We have: \[\rho = \sqrt{x^2 + y^2}\] \[\cos \phi = \frac{x}{\sqrt{x^2 + y^2}}\] \[\sin \phi = \frac{y}{\sqrt{x^2 + y^2}}\] And thus, the Cartesian form of \(\mathbf{H}\) is: \[\mathbf{H} = 3\sqrt{x^2+y^2}\left(-\frac{y}{x^2+y^2}\mathbf{i}+\frac{x}{x^2+y^2}\mathbf{j}\right)\]

Next, we need to calculate the curl of \(\mathbf{H}\). We can do this using the curl formula in Cartesian coordinates: \[\nabla \times \mathbf{H} = \left(\frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z}\right)\mathbf{i} - \left(\frac{\partial H_x}{\partial z} - \frac{\partial H_z}{\partial x}\right)\mathbf{j} + \left(\frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y}\right)\mathbf{k}\] Since there is no \(H_z\) component, the curl simplifies to: \[\nabla \times \mathbf{H} = \left(\frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y}\right)\mathbf{k}\] Now let's compute the partial derivatives. From the expression of \(\mathbf{H}\), we have: \[H_x = -\frac{3xy}{x^2+y^2}\] \[H_y = \frac{3x^2}{x^2+y^2}-3\] Thus: \[\frac{\partial H_y}{\partial x}=\frac{6x^2y}{(x^2+y^2)^2}\] \[\frac{\partial H_x}{\partial y}=\frac{6x^3-6xy^2}{(x^2+y^2)^2}\] Now we can find \(\nabla \times \mathbf{H}\): \[\nabla \times \mathbf{H} = \left(\frac{6x^2y}{(x^2+y^2)^2} - \frac{6x^3-6xy^2}{(x^2+y^2)^2}\right)\mathbf{k} = \frac{6xy}{(x^2+y^2)^2}\mathbf{k}\]

Finally, we need to compute the surface integral of the curl of \(\mathbf{H}\) over the given surface \(S_1\). We will use the following formula: \[\int_{S_1} (\nabla \times \mathbf{H}) \cdot d\mathbf{S}\] As \(S_1\) is the part of the sphere of radius \(r=4\) centered at \((0,0,3)\) that lies above the \(xy\)-plane, the parametrization of \(S_1\) is given by: \[\mathbf{R}(\theta, \phi) = \langle 4 \sin \theta \cos \phi, 4 \sin \theta \sin \phi, 3 + 4 \cos \theta\rangle\] where \(0\le\theta\le\pi/2\) and \(0\le\phi\le2\pi\). In order to compute the surface integral, we need to find \(d\mathbf{S}\). First, we compute the partial derivatives of \(\mathbf{R}\) with respect to \(\theta\) and \(\phi\): \[\frac{\partial\mathbf{R}}{\partial\theta} = \langle 4 \cos \theta \cos \phi, 4 \cos \theta \sin \phi, -4 \sin \theta\rangle\] \[\frac{\partial\mathbf{R}}{\partial\phi} = \langle -4 \sin \theta \sin \phi, 4 \sin \theta \cos \phi, 0\rangle\] Now compute the cross product to find \(d\mathbf{S}\): \[d\mathbf{S}= \frac{\partial\mathbf{R}}{\partial\theta} \times \frac{\partial\mathbf{R}}{\partial\phi} = \langle 16 \sin^2 \theta \cos \phi, 16 \sin^2 \theta \sin \phi, 16 \sin \theta \cos \theta\rangle \, d\theta d\phi\] Since only the \(z\) component of \(\nabla \times \mathbf{H}\) is nonzero, which is the last component of the above vector, so the dot product simplifies to: \[(\nabla \times \mathbf{H}) \cdot d\mathbf{S} = 96 \sin \theta \cos \theta\cdot\frac{6xy}{(x^2+y^2)^2} \, d\theta d\phi\] Now compute the integral: \[\int_{S_1} (\nabla \times \mathbf{H})\cdot d\mathbf{S} = 96\int_{0}^{\pi/2}\int_{0}^{2\pi} \sin \theta \cos \theta\cdot\frac{6xy}{(x^2+y^2)^2} \, d\theta d\phi\] As you note that we have to change \(x\), \(y\) from Cartesian coordinates to its polar coordinates, however after doing so you might note that it is becoming very hard to integrate. Instead of this method try to find the flux of \(\nabla \times \mathbf{H}\) by using Stoke's theorem and thereby seeing that the circuit integral evaluates to \(0\), so the value of flux integral is also \(0\).