Problem 44 - Thomas Calculus

Question

Find the area of the upper portion of the cylinder \(x^{2}+z^{2}=1\) that lies between the planes \(x=\pm 1 / 2\) and \(y=\pm 1 / 2.\)

Step-by-Step Solution

Answer

The area of the upper portion is \(\frac{2\pi}{3}\).

Step 1: Understand the Problem

You are given the equation of a cylinder in the form of \(x^2 + z^2 = 1\). This describes a cylinder with radius 1, centered along the y-axis. You need to find the area of the upper portion of this cylinder that is bounded by the planes \(x=\pm\frac{1}{2}\) and \(y=\pm\frac{1}{2}\).

Step 2: Determine the Area Section on the Cylinder Surface

The surface area of interest is on the cylinder between \(x= -\frac{1}{2}\) and \(x= \frac{1}{2}\), and similarly between \(y= -\frac{1}{2}\) and \(y= \frac{1}{2}\). The portion of the cylinder lying above the plane \(z = 0\) is considered as the upper portion.

Step 3: Set Up the Integral for Surface Area

The general formula for the surface area \(A\) over a surface described by a parameterized coordinate \((x, y, z)\) is given by the expression:\[ A = \iint \sqrt{\left(\frac{\partial x}{\partial u}\right)^2 + \left(\frac{\partial x}{\partial v}\right)^2 + 1}\, du\, dv \]For the cylinder, parameterize the surface with polar coordinates \((r, \theta)\), where \(x = r\cos(\theta), \ z = r\sin(\theta)\) and \(r = 1\).

Step 4: Solve for Surface Area in Given Bounds

To solve for the integral, we note the cylinder's symmetry and set integration bounds for each variable:\(y\) goes from \(-\frac{1}{2}\) to \(\frac{1}{2}\).For \(x = \frac{1}{2}\), we have \( \arccos(\frac{1}{2})\) which bounds \(\theta\) from \(-\frac{\pi}{3}\) to \(\frac{\pi}{3}\).The integral becomes:\[ A = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \int_{-\frac{1}{2}}^{\frac{1}{2}} r \, d\theta \, dy \]Where \(r = 1\).

Step 5: Perform the Integration

Substitute \(r = 1\) into the integral. The integral simplifies to:\[ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \int_{-\frac{1}{2}}^{\frac{1}{2}} dy \, d\theta \]Solve the inner integral for \(y\):\[ \int_{-\frac{1}{2}}^{\frac{1}{2}} dy = 1 \]Then the outer integral becomes:\[ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} d\theta = \frac{2\pi}{3} \]Thus, the total area is \(\frac{2\pi}{3}\).

Key Concepts

Cylinder Geometry

Cylinders are intriguing shapes that you might come across frequently in geometry. Imagine stretching a circle upwards to form a long, tube-like shape. This is what we typically call a cylinder.
For the problem at hand, we are dealing with a cylinder described by the equation \(x^2 + z^2 = 1\). This tells you that:

• The cylinder is centered along the y-axis since the equation contains just \(x\) and \(z\).
• It has a radius of 1.

Picture this as a circular shape when you look from above, with the cylinder extending infinitely along the y-axis in either direction. The slice of the cylinder between \(x=-\frac{1}{2}\) and \(x=\frac{1}{2}\), and similarly between \(y=-\frac{1}{2}\) and \(y=\frac{1}{2}\), is the area of interest for our problem. We focus only on the part of the cylinder that lies above the plane \(z=0\), called the "upper portion."

Surface Integration

Surface integration is like finding how much wrapping paper you need to cover a surface completely. In our case, it's about wrapping the curved surface of a cylinder.
To figure out the surface area, we use a tool from calculus known as surface integration. This helps us "add up" small bits of surface across the entire region of interest.
The general formula for surface area you've come across in the problem is: \[ A = \iint \sqrt{\left(\frac{\partial x}{\partial u}\right)^2 + \left(\frac{\partial x}{\partial v}\right)^2 + 1}\, du\, dv \] This involves math terms like derivatives to deal with the changes along the surface, but don't worry too much about that. What’s happening here is that for each little patch of surface, we calculate how much area it contributes, and then we "sum up" all those contributions via integration.

Parametrization

To make the cylinder easier to work with mathematically, we use a method called parametrization. Think of it like using a map to navigate an unfamiliar area.
In geometry, parametrization means expressing the coordinates (\(x, y, z\)) as functions of one or more variables. For this problem, those variables are \(r\) and \(\theta\) from polar coordinates, where:

• \(x = r \cos(\theta)\)
• \(z = r \sin(\theta)\)
• \(r = 1\) since the radius of the cylinder is 1.

By setting up the problem this way, it becomes significantly easier to calculate the area of the portion we are interested in. We planned our ‘route’ around the cylinder using \(\theta\) and moved vertically using \(y\), which makes integrating possible and direct. The bounds for \(\theta\) result from the geometric constraints, ensuring our integration covers precisely the right section of the cylinder.