Question

Find the area of the part of the cylinder \(y^{2}+z^{2}=4\) in the first octant, cut out by the planes \(x=0\) and \(y=x\).

Short answer

The area is \( \frac{\pi}{2} + 1 \).

Step-by-step solution

- Understand the Problem

The goal is to find the area of a specific part of the surface of a cylinder defined by the equation \(y^{2}+z^{2}=4\), within the constraints of being in the first octant and cut out by the planes \(x=0\) and \(y=x\).

- Identify the bounds

First octant means all coordinates \(x\), \(y\), and \(z\) are non-negative. The plane \(x=0\) is the yz-plane, and the plane \(y=x\) is a diagonal plane intersecting the first octant.

- Parameterize the Cylinder

Parameterize the surface of the cylinder where \(y = 2 \cos(\theta)\) and \(z = 2 \sin(\theta)\). The value of \(\theta\) will range from \(0\) to \(\frac{\pi}{2}\) to stay within the first octant.

- Setup the Integral

The infinitesimal surface area on the cylinder can be given by: \(dA = 2 \cos(\theta) \, d\theta \, dx\). Integrate \(x\) from \(0\) to \(2 \cos(\theta)\) and \(\theta\) from \(0\) to \(\frac{\pi}{4}\).

- Calculate the Integral

We set up the integral: \[A = \int_{0}^{\frac{\pi}{4}} \int_{0}^{2 \cos(\theta)} 2 \cos(\theta) \, dx \, d\theta\]Evaluate the inner integral first with respect to \(x\): \[\int_{0}^{2 \cos(\theta)} 2 \cos(\theta) \, dx = 4 \cos^{2}(\theta)\] Then integrate with respect to \(\theta\): \[A = \int_{0}^{\frac{\pi}{4}} 4 \cos^{2}(\theta) \, d\theta\]. Use the double-angle formula \(\cos^{2}(\theta) = \frac{1 + \cos(2\theta)}{2}\) to simplify: \[A = 2 \int_{0}^{\frac{\pi}{4}} (1 + \cos(2\theta)) \, d\theta\]. Upon integration: \[A = 2 \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_{0}^{\frac{\pi}{4}}\].After substituting the limits and solving, the final area is found.

- Evaluate the Result

Substitute the limits into the antiderivative: \[A = 2 \left( \frac{\pi}{4} + \frac{1}{2} \sin(\frac{\pi}{2}) \right) - 2 \left( 0 + \frac{1}{2} \sin(0) \right)\]. Simplifying gives: \[A = 2 \left( \frac{\pi}{4} + \frac{1}{2} \right)\ = 2 \left( \frac{\pi}{4} + \frac{1}{2} \right)\ = \frac{\pi}{2} + 1\].

Key concepts

surface area integral

To find the area of a surface within a specific region, we use a surface area integral. This type of integral helps us sum up infinitesimally small area elements over the entire region of interest. For the given problem, we seek the area on a cylinder, traditionally described with specific bounds or parameters.
Imagine slicing the cylinder's curved surface into tiny patches and then summing up the area of each patch. That's essentially what a surface area integral does.
The infinitesimal surface area element is often denoted as \(dA\). In this specific problem, it's given by \(dA = 2 \, \text{cos}(\theta) \, d\theta \, dx\).
This formula takes into account the contributions of both \( \theta \) and \( x \), the parameters of the surface part.

• In summary, we break down the surface into manageable components.
• Then, we systematically integrate across the specified bounds.

By carefully setting up and evaluating the surface area integral, we accurately determine the area of the desired surface portion.

parameterization of surfaces

In problems involving integrals over surfaces, it’s crucial to express the surface parametrically. Parameterization involves representing the surface in terms of simpler, alternative variables. This makes complex problems more tractable.
For a cylinder defined by \(y^2 + z^2 = 4\), a convenient parameterization uses angles and distances:
We represent the cylinder’s surface with
\(y = 2 \, \text{cos}(\theta)\) and \(z = 2 \, \text{sin}(\theta)\).
These equations come from trigonometric relationships in the unit circle scaled by radius 2. With \( \theta \) ranging from 0 to \( \frac{\text{π}}{2} \), we cover the first octant.
Additionally, we consider \( x \) going from 0 to the intercepted values on the plane \( y = x \).

• Hence, the parameterization helps us reformulate the problem in terms of \( \theta \) and \( x \).
• This substitution simplifies setting up the integral.

Overall, surface parameterization transforms a seemingly complex surface area calculation into a more straightforward task.

multivariable calculus

Multivariable calculus allows us to handle problems involving more than one independent variable. When finding the surface area of part of a cylinder as described, we’re working with multiple variables like \( x \), \( y \), and \( \theta \).
Here’s a quick breakdown of its importance:

• **Multiple Integration:** Integrating functions over regions defined by more than one variable.
• **Parameterization:** Expressing surfaces using alternative, simpler variables to ease the integration.
• **Handling Constraints:** By working in regions like the first octant or bounded by planes like \( x = 0 \) and \( y = x \).

Each step we take requires a solid understanding of multivariable calculus to meticulously set up the integrals and evaluate them correctly.
In the exercise, integrating over \( x \) and \( \theta \) efficiently gives us the area thanks to fundamental techniques in multivariable calculus.