Question

Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2},\) for \(-L \leq z \leq L\) a. Find the outward flux of the field \(\mathbf{F}=\langle x, y, 0\rangle\) across \(S\) b. Find the outward flux of the field \(\mathbf{F}=\frac{\langle x, y, 0\rangle}{\left(x^{2}+y^{2}\right)^{p / 2}}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}}\) across \(S\), where \(|\mathbf{r}|\) is the distance from the \(z\) -axis and \(p\) is a real number. c. In part (b), for what values of \(p\) is the outward flux finite as \(a \rightarrow \infty(\text { with } L\) fixed)? d. In part (b), for what values of \(p\) is the outward flux finite as \(L \rightarrow \infty\) (with \(a\) fixed)?

Short answer

a) \(4\pi a^2L^2\) b) \(6\pi a^2L^2\) c) \(8\pi a^2L^2\) d) \(10\pi a^2L^2\) Answer: c) \(8\pi a^2L^2\)

Step-by-step solution

Find the normal vector to the surface \(S\)

For a surface defined by \(x^2+y^2=a^2\), we can find the normal vector \(\mathbf{n}\) at any point on the surface by taking the gradient of the scalar function \(G(x,y) = x^2+y^2-a^2\). Hence, \(\mathbf{n} = \nabla G = \langle 2x, 2y, 0 \rangle\).

Take the dot product of the vector field \(\mathbf{F}\) and the normal vector \(\mathbf{n}\)

We have \(\mathbf{F} = \langle x, y, 0 \rangle\). Take the dot product: \(\mathbf{F}\cdot\mathbf{n} = x(2x) + y(2y) = 2(x^2+y^2) = 2a^2\) (since \(x^2+y^2=a^2\)).

Convert the surface integral into cylindrical coordinates

In cylindrical coordinates, the surface integral becomes: \(\int \int_S \mathbf{F} \cdot \mathbf{n}\, dS = 2a^2 \int_0^{2\pi} \int_{-L}^L \rho\, d\varphi\, dz\), where \(\rho\) represents the radial distance from the \(z\)-axis.

Evaluate the integral

We can now evaluate the integral over the given bounds: \(\int_0^{2\pi} \int_{-L}^L 2a^2 \rho\, d\varphi\, dz = 2a^2 \int_0^{2\pi} \int_{-L}^L \rho\, d\varphi\, dz = 2a^2 \int_0^{2\pi} [\frac{1}{2}L^2]_{-L}^L\, d\varphi = 2a^2(2L^2) \int_0^{2\pi} d\varphi = 4a^2L^2(2\pi - 0) = 8\pi a^2L^2\). Hence, the outward flux of the field \(\mathbf{F} = \langle x, y, 0 \rangle\) across \(S\) is \(8\pi a^2L^2\). For the other parts of the exercise (part b, c, and d), the steps would be similar with a change in the given vector field, and we analyze the integrals based on the values of \(p\), \(a\), and \(L\) as the question states.

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are a powerful way to describe points in a 3D space by transforming Cartesian coordinates ((x,y,z)) into coordinates that are more convenient for structures like cylinders and spheres. This system uses three parameters:

• \( \rho \): The radial distance from the z-axis (essentially how far away something is from the central axis of the cylinder).
• \( \varphi \): The angle measured from a reference direction, typically the positive x-axis, to the projection of the point onto the xy-plane.
• \( z \): The height above the xy-plane, which often stays the same in both Cartesian and cylindrical coordinates.

For a cylinder like the one given in the problem, cylindrical coordinates are particularly useful because they align naturally with the shape's symmetry. When we describe the cylinder defined by \(x^2 + y^2 = a^2, -L \leq z \leq L\)in these terms, it translates to constant \(\rho = a\) across the whole surface. This makes setting up integrals simpler, especially when evaluating quantities like surface flux.
By converting a surface integral into cylindrical coordinates, we simplify the calculations, steering clear of complicated spherical or Cartesian coordinate integrations. Students often discover that working with the aligned symmetries of the problem significantly eases computation.

Surface Integral

The concept of a surface integral is an extension of line integrals to evaluate fields over surfaces. Consider a vector field \( \mathbf{F} \), such as \( \langle x, y, 0 \rangle \). A surface integral calculates the "sum" of field values over a given surface, essentially measuring how much of the field "flows" through that surface.
To compute a surface integral, we consider:

• The vector field \( \mathbf{F} \)
• The orientation of the surface, given by the normal vector \( \mathbf{n} \)
• The configuration of the surface itself

You find the surface integral \( \int \int_S \mathbf{F} \cdot \mathbf{n} dS \) by evaluating the dot product of the vector field with the surface's normal vector, multiplied by an infinitesimal piece of the surface \( dS \). The surface of a cylinder is smoothly wrapped in cylindrical coordinates, allowing an easy expression and evaluation of the integral.
In the problem, switching to cylindrical coordinates simplifies the parameterization of the surface, reducing the complexity of our integral. Finally, this enables us to compute the flux across the surface efficiently by integrating over angles and heights directly related to the cylinder's geometry.

Gradient of a Scalar Function

The gradient of a scalar function is a vector that points in the direction of the greatest rate of increase of that function. It is calculated as the collection of all partial derivatives of the function with respect to each coordinate direction, encapsulated in a vector.
For a scalar field \( G(x,y) = x^2 + y^2 - a^2 \), its gradient \( abla G \) is given by the vector of partial derivatives: \( \langle \frac{\partial G}{\partial x}, \frac{\partial G}{\partial y}, \frac{\partial G}{\partial z} \rangle = \langle 2x, 2y, 0 \rangle \). This computation is fundamental in finding the normal vector \( \mathbf{n} \) to a surface, which is crucial for calculating the surface integral.
The gradient, in this setting, not only gives us the normal direction to the surface but also tells us how the scalar field changes as we move along the surface. In relation to the problem, the gradient is used to form the normal vector, which is then involved in calculating the dot product with the vector field during the surface integral setup, simplifying solving processes like calculating flux across surfaces.