Problem 284 - 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. \(\mathbf{F}(x, y, z)=x \mathbf{i}+2 y \mathbf{j}-3 z \mathbf{k}, \quad\) and \(S\) is that part of plane \(15 x-12 y+3 z=6\) that lies above unit square \(0 \leq x \leq 1,0 \leq y \leq 1\)

Step-by-Step Solution

Answer

The value of the surface integral is -18.

Step 1: Determine Surface Equation and Normal Vector

The surface in question is a plane described by the equation \(15x - 12y + 3z = 6\). We can express \(z\) in terms of \(x\) and \(y\) to describe the surface explicitly: \(z = \frac{6 - 15x + 12y}{3} = 2 - 5x + 4y\).The vector normal to the surface, \(\mathbf{N}\), can be found from the coefficients of \(x\), \(y\), and \(z\) in the plane equation, which gives \(\mathbf{N} = \langle 15, -12, 3 \rangle\). This is an outward normal vector.

Step 2: Construct the Surface Integral Expression

The integral \(\iint_{S} \mathbf{F} \cdot \mathbf{N} \, dS\) requires us to evaluate \(\mathbf{F} \cdot \mathbf{N}\) and integrate over the projection of \(S\) onto the \(xy\)-plane. The dot product is:\(\mathbf{F} \cdot \mathbf{N} = (x, 2y, -3z) \cdot (15, -12, 3) = 15x - 24y - 9z\).Substituting \(z = 2 - 5x + 4y\) from the plane equation, we have:\(15x - 24y - 9(2 - 5x + 4y) = 15x - 24y - 18 + 45x - 36y = 60x - 60y - 18\).Therefore, \(\mathbf{F} \cdot \mathbf{N} = 60x - 60y - 18\).

Step 3: Set Up the Double Integral

We need to integrate over the unit square in the \(xy\)-plane, \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\). The integral is:\[ \iint_{0 \leq x \leq 1, \, 0 \leq y \leq 1} (60x - 60y - 18) \, dA \]where \(dA = dx \, dy\).

Step 4: Evaluate the Double Integral

First, integrate with respect to \(y\):\[ \int_{0}^{1} (60x - 60y - 18) \, dy = \left[ 60xy - 30y^2 - 18y \right]_{0}^{1} = 60x - 30 - 18 = 60x - 48 \]Next, integrate the result with respect to \(x\):\[ \int_{0}^{1} (60x - 48) \, dx = \left[ 30x^2 - 48x \right]_{0}^{1} = 30(1)^2 - 48(1) = 30 - 48 = -18 \].

Key Concepts

Surface Integral

A surface integral is an essential concept in vector calculus, allowing you to extend the ideas of line integrals to surfaces. In essence, it provides a way to accumulate a function over a surface rather than along a curve. This comes in handy in fields like physics and engineering, where you often need to calculate quantities distributed over a surface, such as electric flux or fluid flow.

To perform a surface integral, you follow these general steps:

• Define the surface over which you integrate, usually described by a parameterization or an explicit equation like in this exercise.
• Identify the vector field being integrated over the surface.
• Calculate the dot product of the vector field and the normal vector to the surface.
• Integrate this dot product over the entire surface.

In the problem given, the surface is defined by the plane equation, and we want to integrate the vector field over this plane's section using the formula for surface integrals.

Dot Product

The dot product, also called the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (typically vectors) and returns a single number. The importance of the dot product in the context of surface integrals lies in its ability to relate vector fields and surface normals.

The dot product is defined as follows for two vectors \( \mathbf{A} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{B} = \langle b_1, b_2, b_3 \rangle \):

• \( \mathbf{A} \cdot \mathbf{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

In the exercise, you calculated \( \mathbf{F} \cdot \mathbf{N} \), where \( \mathbf{F} \) is the vector field given by \( x \mathbf{i} + 2y \mathbf{j} - 3z \mathbf{k} \) and \( \mathbf{N} \) is the normal vector to the surface, \( \langle 15, -12, 3 \rangle \). The resulting expression is integrated over the surface.

By calculating the dot product, you are essentially capturing the component of the vector field that is parallel to the normal at each point on the surface.

Double Integral

The double integral is used to compute the integration of functions of two variables over a region in the \(xy\)-plane. In our context, it's utilized to evaluate the surface integral by reducing the problem to an integral over a simpler region, often a rectangle in the plane.

Performing a double integral typically involves:

• Setting the integration limits for both variables. For this problem, the integration limits are from 0 to 1 for both \(x\) and \(y\), forming a unit square.
• Evaluating the integral sequentially over one variable and then the other. In this exercise, this involves first integrating with respect to \(y\) and then \(x\).

The double integral \[\iint (60x - 60y - 18) \ dydx\]represents the integral of the dot product over the entire unit square. It simplifies the complex surface integration to something manageable.

Normal Vector

A normal vector to a surface is a vector that is perpendicular to the surface at a given point. It's crucial in calculating surface integrals because it helps in projecting the vector field onto the surface.

For any surface in three-dimensional space described by \(ax + by + cz = d\), the coefficients \(a, b,\) and \(c\) can be used directly to form the normal vector \( \mathbf{N} = \langle a, b, c \rangle \).

The role of the normal vector is profound in surface integrals. It ensures that we measure the appropriate component of the vector field that flows through the surface. The normal vector acts like a scale that turns directional information into a height that's integrated across the plane.

In the problem presented, the normal vector \( \mathbf{N} = \langle 15, -12, 3 \rangle \) directly uses the coefficients from the plane equation, providing a consistent outward direction as required for our surface integral.