Problem 6 - Thomas Calculus

Question

In Exercises \(1-8,\) integrate the given function over the given surface. Cone \(\quad F(x, y, z)=z-x, \quad\) over \(\quad\) the cone \(\quad z=\sqrt{x^{2}+y^{2}}\) \(0 \leq z \leq 1\)

Step-by-Step Solution

Answer

The surface integral over the cone is \( \frac{2\pi}{3} \).

Step 1: Describe the Surface

The surface of integration is a cone given by the equation \( z = \sqrt{x^2 + y^2} \). This describes a cone that is symmetric around the z-axis and extends from the apex at the origin (0,0,0) upward, and we integrate over this surface from \( z = 0 \) to \( z = 1 \).

Step 2: Set Up the Surface Integral

The vector field \( F(x, y, z) = z - x \) is given, and we need to integrate this over the cone surface. A parametrization for the cone surface can be given by cylindrical coordinates: \( x = r\cos(\theta) \), \( y = r\sin(\theta) \), and \( z = r \), with \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq 2\pi \).

Step 3: Compute the Surface Element

The differential surface element is obtained by using the position vector \( \oldsymbol{r}(r, \theta) = (r\cos(\theta), r\sin(\theta), r) \). Its partial derivatives are \( \oldsymbol{r}_r = (\cos(\theta), \sin(\theta), 1) \) and \( \oldsymbol{r}_{\theta} = (-r\sin(\theta), r\cos(\theta), 0) \). The cross-product \( \oldsymbol{r}_r \times \oldsymbol{r}_{\theta} \) gives the normal vector, which yields \( \vec{n}\,dS = r\, dr\, d\theta \).

Step 4: Evaluate the Integral

Substitute the parameters into the function: \( F(x, y, z) = r - r\cos(\theta) \). The integral becomes \( \int_0^{2\pi} \int_0^1 (r - r\cos(\theta))\, r\, dr\, d\theta \). Simplifying this gives \( \int_0^{2\pi} \int_0^1 (r^2 - r^2 \cos(\theta))\, dr\, d\theta \).

Step 5: Integrate with Respect to \( r \)

Evaluate \( \int_0^1 (r^2 - r^2 \cos(\theta))\, dr \). This integral is linear, yielding \( \left[ \frac{r^3}{3} - \frac{r^3}{3} \cos(\theta) \right]_0^1 = \frac{1}{3}(1 - \cos(\theta)) \).

Step 6: Integrate with Respect to \( \theta \)

Now, integrate with respect to \( \theta \): \( \int_0^{2\pi} \frac{1}{3}(1 - \cos(\theta))\, d\theta = \frac{1}{3} \left[ \theta - \sin(\theta) \right]_0^{2\pi} \). Evaluate to get \( \frac{1}{3} (2\pi) = \frac{2\pi}{3} \) since the integral of \( \sin(\theta) \) over \( [0, 2\pi] \) is zero.

Key Concepts

Cone Surface

A cone surface is a three-dimensional geometric figure characterized by its circular base and pointed apex. The particular cone given in this exercise is described by the equation \( z = \sqrt{x^2 + y^2} \). This represents a right circular cone, symmetric around the z-axis. The apex is at the origin (0,0,0), and the cone extends upwards. For integration purposes, we consider the portion of the cone where the height \( z \) ranges from 0 to 1. This specifies a vertical slice of the cone, creating a finite cap on this otherwise infinite shape.

Cylindrical Coordinates

Cylindrical coordinates are extremely useful in problems involving symmetrical shapes like cones or cylinders. They provide a simple way to express points in 3D space using a radius \( r \), an angle \( \theta \), and a height \( z \). For the cone in this exercise, we use the transformation:

• \( x = r\cos(\theta) \)
• \( y = r\sin(\theta) \)
• \( z = r \)

The variables \( r \) and \( \theta \) make handling rotational symmetry straightforward. This transformation simplifies the process of solving the surface integral over a cone, reducing a complex problem into a more manageable one by limiting the z-values from 0 to 1 and \( \theta \) from 0 to \( 2\pi \).

Parametrization

Parametrization involves expressing a surface using a set of parameters, thereby converting a 3D surface into a more tractable form. For the given cone surface, this technique involves re-expressing its coordinates through the parameters \( r \) and \( \theta \). The position vector \( \mathbf{r}(r, \theta) = (r\cos(\theta), r\sin(\theta), r) \) is used.

• \( \mathbf{r}_r = (\cos(\theta), \sin(\theta), 1) \)
• \( \mathbf{r}_\theta = (-r\sin(\theta), r\cos(\theta), 0) \)

These derivative vectors are then used to compute the surface element by taking their cross-product. Parametrization is essential as it converts the problem of integrating over a surface to more familiar double integrals over the parameters.

Surface Element

The surface element, denoted as \( dS \), is a small "piece" of the surface area over which integration is performed. In our problem, after parametrization, we calculate the differential area element using the cross-product of the partial derivatives of the position vector. Specifically, the normal vector comes from \( \mathbf{r}_r \times \mathbf{r}_\theta \).

• This results in \( \vec{n} dS = r dr d\theta \).

In essence, \( dS \) provides the magnitude of the surface area to be considered at each point \( (r, \theta) \) on the parameterized cone. This element is crucial in setting up the surface integral, ensuring that integration is done accurately over the specified region of the surface.