Problem 40 - Advanced Engineering Mathematics

Question

If \(\sigma(x, y, z)\) is charge density in an electrostatic field, then the total charge on a surface \(S\) is \(Q=\iint_{S} \sigma(x, y, z) d S\). Find the total charge on that part of the hemisphere \(z=\sqrt{16-x^{2}-y^{2}}\) that is inside the cylinder \(x^{2}+y^{2}=9\) if the charge density at a point \(P\) on the surface is directly proportional to distance from the \(x y\) -plane.

Step-by-Step Solution

Answer

The total charge is given by \(Q = 36\pi k\).

Step 1: Understanding the Problem

We need to compute the total charge over a hemisphere given by the equation \(z=\sqrt{16-x^{2}-y^{2}}\), but only the part inside the cylinder described by \(x^{2}+y^{2}=9\). The charge density is proportional to the distance from the \(xy\)-plane, which implies \(\sigma(x, y, z) = k \cdot z\), where \(k\) is a constant.

Step 2: Setting up the Integral

The total charge \(Q\) is computed using the surface integral \(Q = \iint_{S} \sigma(x, y, z) \ dS\). Substituting the charge density, we get \(Q = \iint_{S} k \cdot z \ dS\). Since \(z = \sqrt{16-x^{2}-y^{2}}\) describes the upper hemisphere, \(z\) can be expressed in terms of \(x\) and \(y\).

Step 3: Converting to Polar Coordinates

To simplify the integration, we switch to polar coordinates: \(x = r \cos\theta\), \(y = r \sin\theta\), and hence \(z = \sqrt{16 - r^2}\). This transforms the problem into \(Q = \int_{0}^{2\pi} \int_{0}^{3} k \sqrt{16 - r^2} \cdot r \, dr \, d\theta\), because the hemisphere is bounded by the cylinder with \(r \leq 3\).

Step 4: Computing the Surface Area Element

The differential surface area element \(dS\) for the hemisphere can be found by parameterizing the surface and using the equation \(dS = \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2} \, dr \, d\theta\). Around this region, \(\frac{\partial z}{\partial x} = -\frac{x}{\sqrt{16-x^2-y^2}}\) and \(\frac{\partial z}{\partial y} = -\frac{y}{\sqrt{16-x^2-y^2}}\), leading to \(dS = \frac{4}{\sqrt{16-x^2-y^2}} \, r \, dr \, d\theta\) after some algebraic simplification.

Step 5: Solving the Integral

Substitute \(z = \sqrt{16-r^2}\) and \(dS = \frac{4}{\sqrt{16-r^2}} \, r \, dr \, d\theta\) into the integral: \(Q = \int_{0}^{2\pi} \int_{0}^{3} k \sqrt{16 - r^2} \cdot \frac{4r}{\sqrt{16 - r^2}} \, dr \, d\theta\). Simplifying, this becomes \(Q = 4k \int_{0}^{2\pi} \int_{0}^{3} r \, dr \, d\theta\).

Step 6: Evaluating the Integral

The inner integral evaluates to \(\int_{0}^{3} r \ dr = \left[\frac{r^2}{2}\right]_{0}^{3} = \frac{9}{2}\). The outer integral evaluates to \(\int_{0}^{2\pi} \, d\theta = 2\pi\). Thus, the total charge is \(Q = 4k \cdot \frac{9}{2} \cdot 2\pi = 36\pi k\).

Key Concepts

Charge Density

In the context of surface integrals, **charge density** can be thought of as the concentration of electric charge per unit area on a surface. For a surface in an electrostatic field, the charge density \( \sigma(x, y, z) \) is often a function of the coordinates of the surface. It tells us how much charge is present at each point. In this exercise, the charge density is proportional to the distance from the xy-plane. This means that the further away a point P is from the plane, the higher the charge density at that point. This relationship is expressed mathematically as \( \sigma(x, y, z) = k \cdot z \), where k is a proportionality constant and z represents the vertical distance from the xy-plane. To compute the total charge on a part of the surface, it's necessary to integrate the charge density over the entire surface area of interest.

Electrostatic Field

An **electrostatic field** is a field produced by stationary electric charges. It represents the force that a charge would experience if it were placed within the field. In the context of this problem, the charge density is affected by its position in the electrostatic field, but the problem simplifies to finding the total charge on the surface rather than analyzing the field's vector properties. However, understanding electrostatic fields helps us recognize why charge might distribute unevenly over surfaces, often influenced by geometric and spatial factors. Here, since charge density is proportional to the z-coordinate, the electrostatic field's effect on more charged areas may differ from less charged ones, representing this simple relationship in a mathematical form.

Polar Coordinates

To solve the problem of integrating over a hemisphere, converting to **polar coordinates** simplifies the process. Polar coordinates are often preferred when dealing with problems involving circular symmetry, as they simplify boundaries and calculations. In polar coordinates, the position of a point is given by \( (r, \theta) \), where \( r \) is the radial distance from the origin, and \( \theta \) is the angle from the positive x-axis. For this problem, we denote the transformations as: \( x = r \cos\theta \) and \( y = r \sin\theta \). The original hemisphere described by \( z=\sqrt{16-x^{2}-y^{2}} \) becomes \( z=\sqrt{16-r^{2}} \) in polar coordinates. This makes integration over regions like the bounds of a cylinder, \( 0 \leq r \leq 3 \), much more straightforward.

Surface Area Element

The **surface area element** \( dS \) is a small piece of surface area over which the integration is performed. For curved surfaces, calculating the precise form of \( dS \) involves the geometric properties of the surface. In this exercise, we parameterized the hemisphere, obtaining derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) to compute the area element. The essential step is finding the magnitude of the gradient (slope) of the surface, leading to the formula \[dS = \frac{4}{\sqrt{16-x^2-y^2}} \, r \, dr \, d\theta\] after simplification. This formula expresses how much area corresponds to infinitesimal changes in r and \( \theta \), and, when integrated, gives the total area over which to integrate the charge density.