Problem 286 - Calculus Volume 3

Question

For the following exercises, evaluate \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d s\) for vector field \(\mathbf{F}\), where \(\mathbf{N}\) is an outward normal vector to surface S. \(\mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k},\) and \(S\) is the portion of plane \(z=y+1\) that lies inside cylinder \(x^{2}+y^{2}=1\).

Step-by-Step Solution

Answer

The surface integral evaluates to \(\pi\).

Step 1: Understand the Surface and Region

We're evaluating a surface integral over the plane \(z = y + 1\), constrained within the boundary of a cylinder \(x^2 + y^2 = 1\). This means the surface \(S\) lies entirely above this circular base on the \(xy\)-plane.

Step 2: Parameterize the Surface

Since \(z = y + 1\), we can use the parameterization \(\mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} + (y + 1) \mathbf{k}\). This surface is defined within \(-1 \leq x^2 + y^2 \leq 1\).

Step 3: Calculate the Normal Vector

To find an outward normal vector \(\mathbf{N}\), we use the cross product of partial derivatives: \(\mathbf{r}_x = \mathbf{i}\) and \(\mathbf{r}_y = \mathbf{j} + \mathbf{k}\). \(\mathbf{N} = \mathbf{r}_x \times \mathbf{r}_y = \mathbf{-i} + \mathbf{k} \).

Step 4: Form the Dot Product

Compute the dot product \(\mathbf{F} \cdot \mathbf{N}\): \((x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k}) \cdot (-\mathbf{i} + \mathbf{k}) = -x^2 + z^2 = -x^2 + (y+1)^2\).

Step 5: Setup the Surface Integral

Integrate \(-x^2 + (y+1)^2\) over the circle \(x^2 + y^2 \leq 1\). Convert to polar coordinates: \(x = r \cos \theta\), \(y = r \sin \theta\). The bounds are for \(r\) from 0 to 1 and \(\theta\) from 0 to \(2\pi\).

Step 6: Integrate in Polar Coordinates

The integral becomes: \(\int_{0}^{2\pi} \int_{0}^{1} \left( -r^2 \cos^2 \theta + (r \sin \theta + 1)^2 \right) r \, dr \, d\theta\). Solve this integral step-by-step by first integrating over \(r\) and then over \(\theta\).

Step 7: Evaluate the Result

Evaluating the integral yields: first find antiderivatives with respect to \(r\), and evaluate from 0 to 1; then integrate the resulting expression with respect to \(\theta\). Compute the result to find the final value of the surface integral.

Key Concepts

Vector Fields

A vector field is an assignment of a vector to each point in a subset of space. Imagine fields like a breeze through your hair, with each breeze represented as a vector pointing in a specific direction with a certain magnitude. For example, in this exercise, the vector field is given by \( \mathbf{F}(x, y, z) = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \). Here, each component of the vector field (\( x^2 \), \( y^2 \), and \( z^2 \)) is dependent on the respective coordinate, creating direction and magnitude at every point.

• \( \mathbf{i} \) component: changes with \( x^2 \).
• \( \mathbf{j} \) component: changes with \( y^2 \).
• \( \mathbf{k} \) component: changes with \( z^2 \).

Understanding vector fields is crucial because they model real-world phenomena like fluid flow, magnetic fields, and more.

Normal Vector

A normal vector is perpendicular to the surface at a given point. It can be thought of as a spike sticking out of a surface, pointing directly away. For the plane \( z = y + 1 \) in this exercise, determining the normal vector involves using partial derivatives and the cross product. We obtain partial derivatives \( \mathbf{r}_x \) as \( \mathbf{i} \) and \( \mathbf{r}_y \) as \( \mathbf{j} + \mathbf{k} \), which help describe the orientation of the plane. The cross product \( \mathbf{N} = \mathbf{r}_x \times \mathbf{r}_y = -\mathbf{i} + \mathbf{k} \) provides the outward normal vector for the surface. It’s essential in surface integrals, as it determines the direction we measure the vector field across the surface.

Parameterization

Parameterization is like creating a recipe for constructing a surface by switching from usual coordinates (like \( x, y, z \)) into simpler, often easier-to-use coordinates (like \( u, v \)). For the plane \( z = y + 1 \), we use the parameterization form \( \mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} + (y + 1) \mathbf{k} \).

• This transformation helps because it restricts \( S \) inside the cylindrical region visually and mathematically.
• Using parameterization simplifies the complexity of integrating over curves and surfaces.
• Make sure to correctly interpret these parameters, as they dictate the range and limits of the region you will integrate over.

Through parametric equations, integration becomes more approachable since they adapt the surface equations into a more tractable form.

Polar Coordinates

Polar coordinates provide a way to express points on a plane using a radius and an angle, \( (r, \theta) \), compared to Cartesian coordinates \( (x, y) \). This form is especially useful in situations involving circular or rotational symmetry, like our cylinder.In this exercise, we convert the limits of integration for the circle \( x^2 + y^2 = 1 \) into polar coordinates:- \( x = r \cos \theta \)- \( y = r \sin \theta \)These transformations simplify surface integrals over circular areas, turning the complex \( xy \)-plane shapes into simpler \( r \)-based limits, ranging from \( r = 0 \) to \( r = 1 \), and angles \( \theta \) from 0 to \( 2\pi \).This step makes integration much more convenient, as functions involving trigonometric identities often become easier to handle and evaluate.