Problem 10 - Advanced Engineering Mathematics

Question

Find the surface area of the portions of the cone \(z^{2}=\frac{1}{4}\left(x^{2}+y^{2}\right)\) that are within the cylinder \((x-1)^{2}+y^{2}=1\).

Step-by-Step Solution

Answer

The surface area is \( \frac{\sqrt{5} \pi}{2} \).

Step 1: Identifying the Cone

The given equation of the cone is \( z^2 = \frac{1}{4}(x^2 + y^2) \). We can rewrite it in standard form \( z = \pm \frac{1}{2}\sqrt{x^2 + y^2} \), which represents a double cone with its vertex at the origin.

Step 2: Understanding the Cylinder

The equation of the cylinder is \((x-1)^2 + y^2 = 1\), which is a circular cylinder centered at \((1, 0)\) with radius 1. We need to find the intersection of this cylinder with the cone.

Step 3: Setting up the Integration Limits

We need to determine where the cylinder intersects the cone. In the x-y plane, the cylinder's center is at \((1,0)\). To find the actual limits of integration, consider the projection in the x-y plane, transformed by the positions x = x' + 1, y = y': \((x'-1)^2 + y'^2 \leq 1\). This allows us to integrate over the portion defined by the cylinder.

Step 4: Calculating the Surface Element

For a surface given in cylindrical coordinates \(x = r\cos\theta, y = r\sin\theta, z(r) = \frac{1}{2}r\), we use the formula for surface area with parametric surface: \[ A = \int \int \sqrt{x_u^2 + x_v^2 + y_u^2 + y_v^2 + z_u^2 + z_v^2} \, du \, dv \]. Substitute the partial derivatives of \(x, y, z\) with respect to suitable parameters (like \(r\) and \(\theta\)).

Step 5: Solving the Integral

First use \(r\) and \(\theta\) for polar coordinates. Then, note that the angle \(\theta\) bounds should be \(0\) to \(2\pi\), and \(r\) is \(0\) to \(r' = 1 + \cos\theta\). Explicitly evaluate the integral: \[ A = \int_0^{2\pi} \int_0^{1 + \cos\theta} \sqrt{(1 + (1/4))} \, r \, dr \, d\theta \], which simplifies to \( \sqrt{5/4} = \sqrt{5}/2 \) multipliers outside the integral.

Step 6: Integration Result

Complete integration by solving \( \int_0^{2\pi} \int_0^{1+\cos\theta} \frac{\sqrt{5}}{2} r \, dr \, d\theta \). Use fundamental calculus to find the value \(\frac{\sqrt{5}}{2} \int_0^{2\pi} ((1+\cos\theta)^2/2) \, d\theta \), giving rise to a manageable result for surface area.

Key Concepts

Integration Limits

To find the surface area of the conical section inside the cylinder, setting up the correct integration limits is crucial. Integration limits help us determine the region over which we need to integrate, essentially defining the section of the cone within the cylinder.

For this problem, the cylinder is centered at \( (1,0) \) with a radius of 1. This means that in the x-y plane, all points on the surface of the cone must also satisfy the cylinder's equation: \( (x-1)^2 + y^2 = 1 \).

• First, substitute variables to shift the origin to the center of the cylinder: \( x = x' + 1 \) and \( y = y' \).
• This transformation simplifies our region of integration to a standard circle \( x'^2 + y'^2 = 1 \), making it easier to work with.

Next, convert these limits into polar coordinates because they simplify calculations for cylindrical shapes.

• The new integration bounds in polar coordinates (r, θ) become \( 0 \leq r \leq 1 + \, \cos \, \theta \) for the radius, and \( 0 \leq \theta \leq 2\pi \) for the angle.

Cylindrical Coordinates

Cylindrical coordinates are particularly useful when dealing with problems that involve cylindrical or circular symmetry, like in this problem. They use three variables \( (r, \theta, z) \) instead of the Cartesian coordinates \( (x, y, z) \).

• Radius (r): The distance from the point to the z-axis. As the radius increases, the point moves further from the axis.
• Angle (θ): The angle measured counterclockwise from the positive x-axis.
• Height (z): The same as in Cartesian coordinates, representing the vertical position.

For the cone, the equation \( z = \frac{1}{2} r \) reflects the relation between \( z \) and \( r \). Expressing the cone in cylindrical coordinates makes it much easier to work with, especially when integrated over circular or cylindrical regions.

When converting from Cartesian coordinates:

• \( x = r\cos\theta \)
• \( y = r\sin\theta \)
• The z-coordinate remains as given.

Parametric Surface

For the surface area calculation, we often use a parametric representation of the surface. A parametric surface is described using parameters rather than equations in standard form.

For our cone, the parametric surface representation links each point on the cone to a set of (r, θ) values:

• \( x(r, \theta) = r\cos\theta \)
• \( y(r, \theta) = r\sin\theta \)
• \( z(r) = \frac{1}{2} r \)

This representation is very helpful for integrating surfaces, as it allows you to find the surface area directly.

To compute the surface area, we use the formula:\[A = \int \int \sqrt{x_u^2 + x_v^2 + y_u^2 + y_v^2 + z_u^2 + z_v^2} \, du \, dv\]
Here, \( x_u, y_u, z_u \) and \( x_v, y_v, z_v \) are partial derivatives of the parametric equations with respect to parameters, typically (r, θ) in this setting. Such formulas simplify the complex task of finding areas on 3D surfaces.