Problem 315 - Calculus Volume 3

Question

Evaluate surface integral I\int \(y z d\) S, where \(S\) is the part of plane \(z=y+3\) that lies inside cylinder \(x^{2}+y^{2}=1\)

Step-by-Step Solution

Answer

The surface integral evaluates to 0.

Step 1: Parametrize the Surface

We start by parametrizing the surface \( S \), which is the plane \( z = y + 3 \). Since the cylinder \( x^2 + y^2 = 1 \) restricts \( x \) and \( y \), we can use cylindrical coordinates: \( x = \cos \theta \), \( y = \sin \theta \), and \( z = \sin \theta + 3 \). This leads to the parameterization: \( \vec{r}(\theta, y) = (\cos \theta, y, y + 3) \) with restrictions \( -1 \leq y \leq 1 \) and \( 0 \leq \theta \leq 2\pi \).

Step 2: Compute the Differential Element

To find \( dS \), we compute the cross product of partial derivatives of \( \vec{r}(\theta, y) \).\[ \frac{\partial \vec{r}}{\partial \theta} = (-\sin \theta, 0, 0) \quad \text{and} \quad \frac{\partial \vec{r}}{\partial y} = (0, 1, 1) \]Then, the cross product is \( \vec{r}_y \times \vec{r}_\theta = (-\sin \theta, 0, 0) \times (0, 1, 1) = (0, \sin \theta, -\sin \theta) \). The magnitude of this vector gives \( dS = \sqrt{0 + \sin^2 \theta + \sin^2 \theta} = \sin \theta \sqrt{2} \).

Step 3: Set Up the Integrand

The integrand is \( yz \cdot dS \) where \( z = y + 3 \). Substitute the parameterization into the expression, giving \( yz = y(y+3) = y^2 + 3y \). Thus, the integrand becomes \( (y^2 + 3y) \cdot \sin \theta \sqrt{2} \).

Step 4: Evaluate the Double Integral

Now, we integrate over the restricted region: Integrate the expression \((y^2 + 3y)\sin \theta \sqrt{2}\) with respect to \( y \) from \( -1 \) to \( 1 \), then with respect to \( \theta \) from \( 0 \) to \( 2\pi \):\[ I = \int_0^{2\pi} \int_{-1}^{1} (y^2 + 3y) \sin \theta \sqrt{2} \: dy \, d\theta \]First, integrate \( \int_{-1}^{1} (y^2 + 3y) \: dy \). Since the function is odd over the symmetric interval, this integral evaluates to 0. Thus, the entire surface integral is 0.

Key Concepts

Parametrization of Surfaces

Understanding the parametrization of surfaces is crucial in evaluating surface integrals. In this problem, the surface is defined by the plane equation \(z = y + 3\).
To parameterize the given surface, we employ cylindrical coordinates that beautifully handle circular restrictions.

• The cylinder \(x^2 + y^2 = 1\) guides our choice. Thus, we use \(x = \cos \theta\) and \(y = \sin \theta\) to respect the circular symmetry.
• The plane equation informs us that \(z = y + 3\), matching with the cylindrical coordinate selection.

By combining these, we form the parametric representation: \(\vec{r}(\theta, y) = (\cos \theta, y, y + 3)\). This parameterizes the surface within the cylinder, with \(-1 \leq y \leq 1 \) and \(0 \leq \theta \leq 2\pi\).
Parametrization simplifies complex surfaces into two variables, easily manipulated in integrals.

Cylindrical Coordinates

Cylindrical coordinates are an effective tool in examining surfaces like planes within cylindrical boundaries. They bridge the gap between two-dimensional constraints and three-dimensional applications, making calculations more tractable.
Here's how:

• It uses three parameters: \(r\), \(\theta\) (a circular angle), and \(z\). In this context, however, \(r\) is implicitly set by the plane's intersection with the cylinder.
• \(x = \cos \theta\) and \(y = \sin \theta\) satisfy the circular boundary of \(x^2 + y^2 = 1\).
• In essence, cylindrical coordinates allow us to describe surfaces using simpler, intuitive units.

Choosing cylindrical coordinates aligns with our problem's symmetry and constraints, making subsequent mathematical processes seamless.

Vector Calculus

Vector calculus arms us with methods to navigate and analyze vector fields over surfaces. In this problem, it aids with calculations of the differential elements needed for the surface integral.
Critical in this domain is the cross product:

• The partial derivatives of the parametrization \(\vec{r}(\theta, y) = (\cos \theta, y, y + 3)\) are computed first.
• \(\frac{\partial \vec{r}}{\partial \theta} = (-\sin \theta, 0, 0)\) and \(\frac{\partial \vec{r}}{\partial y} = (0, 1, 1)\).
• The cross product \((0, \sin \theta, -\sin \theta)\) emerges from these derivatives, crucial in finding \(dS\).

Such manipulations in vector calculus allow us to unravel the intricacies of the integral, ultimately simplifying its evaluation.

Differential Elements

A differential element, such as \(dS\) on a surface, is fundamental to evaluating surface integrals. These elements capture tiny segments of the surface area and adapt to the geometry of the surface.
In this scenario:

• After obtaining the cross product \((0, \sin \theta, -\sin \theta)\), we determined its magnitude, which yielded \(dS = \sin \theta \sqrt{2}\).
• This magnitude reflects the infinitesimal area, representing a piece of the entire surface within the limits set by the parameters.

Calculating \(dS\) involves understanding the interplay of vector derivatives and geometry. Its role is pivotal as it scales the integrand in response to the surface's spatial orientation and extent.
Knowing these essentials empowers us to tackle broader classes of physical and mathematical problems involving surfaces.