Question

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.

Short answer

Based on the step by step solution, find the flux of the curl of the vector field H through the spherical surface S1 lying above the xy-plane. Answer: The flux of the curl of the vector field H through the spherical surface S1 is 0.

Step-by-step solution

Convert the vector field from cylindrical to Cartesian 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: \[\mathbf{H} = 3\sqrt{x^2+y^2}\left(-\frac{y}{x^2+y^2}\mathbf{i}+\frac{x}{x^2+y^2}\mathbf{j}\right)\]

Calculate the curl of H in Cartesian coordinates

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 \(\nabla \times \mathbf{H}\): \[\nabla \times \mathbf{H} = \left(\frac{6x^2y}{(x^2+y^2)^2} - \frac{6x^3-6xy^2}{(x^2+y^2)^2}\right)\mathbf{k} = \frac{6xy}{(x^2+y^2)^2}\mathbf{k}\]

Compute the surface integral of the curl of H

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 \(0\).

Key concepts

Cylindrical coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height (z-coordinate). They are particularly useful for systems with symmetry around a central axis, such as a cylinder or a sphere with rotation symmetry.
In this system, a point is expressed as \(\mathbf{P} = (\rho, \phi, z)\), where:

• \(\rho\) is the radial distance from the origin to the point's projection in the xy-plane.
• \(\phi\) is the angular coordinate, or azimuth angle, measured from the positive x-axis, similar to the angle in polar coordinates.
• \(z\) is the height above the xy-plane, essentially representing the third dimension.

Using cylindrical coordinates can simplify complex problems, especially when dealing with circular symmetry, as you can align the coordinate system to best fit the problem's geometry.

Cartesian coordinates

Often we use Cartesian coordinates, denoted by \(\mathbf{P} = (x, y, z)\), which are crucial for understanding geometry and performing vector operations in a three-dimensional space. This coordinate system uses perpendicular axes (x-axis, y-axis, and z-axis), and each point in space is defined by its distance along each axis.
Let's see how cylindrical coordinates can be converted into Cartesian coordinates:

• The x-coordinate is determined by \(x = \rho \cos \phi\), representing the horizontal component.
• The y-coordinate, \(y = \rho \sin \phi\), represents the vertical component in the plane.
• The z-coordinate remains unchanged, \(z = z\).

This conversion is vital in fields like electromagnetics, where it's important to express vector fields in different coordinate systems to simplify calculations and physical interpretations.

Surface integral

A surface integral extends the concept of an integral over a curve to a two-dimensional surface. It calculates the sum of a field over a surface area, which can be critical for understanding flux through a surface. Surface integrals are especially important in physics, where they are used to compute quantities like electric flux and magnetic flux.
To perform a surface integral, you need:

• A vector field, e.g., \(abla \times \mathbf{H}\), which you want to integrate over the surface.
• The surface itself, parametrize with coordinates, often given in terms of parameters like \(\theta\) and \(\phi\) for spherical surfaces.
• The differential area element, \(d\mathbf{S}\), which involves taking the cross product of partial derivatives from your parametrization.

By integrating the dot product of the vector field and the surface element over the entire surface, we obtain a total flux, which is incredibly valuable for various applications in electromagnetism and fluid dynamics.

Stoke's theorem

Stoke's theorem is a powerful tool in vector calculus that connects surface integrals to line integrals. It states that the surface integral of the curl of a vector field \(\mathbf{H}\) over a surface \(S\) is equal to the line integral of the vector field along the boundary of the surface.
Mathematically, this is expressed as:

• \(\int_{S} (abla \times \mathbf{H}) \cdot d\mathbf{S} = \oint_{\partial S} \mathbf{H} \cdot d\mathbf{r}\)

Stoke's theorem can simplify problems by reducing a complex surface integral to a more manageable line integral, saving significant computational effort. It also gives insight into the behavior of fields, showing how circulation along a boundary relates to a field's curl within an area.
This theorem is especially useful when evaluating fields around objects like spheres or cylinders, as it can quickly show if the resulting integral is zero, indicating no net flux through the surface.