Problem 30 - Thomas Calculus

Question

In Exercises 29 and 30 , find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. $$\mathbf{F}(x, y, z)=y x^{2} \mathbf{i}-2 \mathbf{j}+x z \mathbf{k}$$ S: rectangular surface \(y=0, \quad-1 \leq x \leq 2, \quad 2 \leq z \leq 7\) direction \(-\mathbf{j}\)

Step-by-Step Solution

Answer

The surface integral is 30.

Step 1: Express Surface Parameters

Given the surface is rectangular and lies in the plane \(y=0\), it is defined by coordinates \(-1 \leq x \leq 2\) and \(2 \leq z \leq 7\). The vector normal to this surface in the direction \(-\mathbf{j}\) is \(-\mathbf{j}\).

Step 2: Define the Surface Element

Because the surface is a flat rectangle on the plane \(y=0\), the differential area vector \(d\mathbf{S}\) is \(dx \, dz \, (-\mathbf{j}) = -dx \, dz \, \mathbf{j}\).

Step 3: Calculate the Dot Product

The dot product \(\mathbf{F} \cdot d\mathbf{S}\) is \([yx^2\mathbf{i} - 2\mathbf{j} + xz\mathbf{k}] \cdot [-dx \, dz \, \mathbf{j}] = 2dx \, dz\). The terms involving \(\mathbf{i}\) and \(\mathbf{k}\) vanish due to orthogonality.

Step 4: Set Up the Double Integral

The surface integral becomes a double integral \(\iint_S 2 \, dA\), where \(dA = dx \, dz\). The limits of integration for \(x\) are from \(-1\) to \(2\), and for \(z\), from \(2\) to \(7\).

Step 5: Solve the Integral

Perform the double integral: \[\int_{2}^{7} \int_{-1}^{2} 2 \, dx \, dz \]. The integral of \(2\) with respect to \(x\) from \(-1\) to \(2\) is \(2[x]_{-1}^{2} = 6\). Then integrate with respect to \(z\): \[\int_{2}^{7} 6 \, dz = 6[z]_{2}^{7} = 30\].

Key Concepts

vector fields

A vector field is a mathematical concept where each point in a space is assigned a vector. Imagine a map of arrows, where each arrow represents a force or direction of influence at that particular point. It is a bit like wind maps on weather reports, where arrows show the direction and strength of the wind at various locations. In this exercise, the vector field is given by \( \mathbf{F}(x, y, z) = y x^{2} \mathbf{i} - 2 \mathbf{j} + x z \mathbf{k} \). Here, \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the standard unit vectors in the x, y, and z direction respectively. Each term in the vector field expression represents a different component or direction in the field.

• \( y x^{2} \mathbf{i} \) indicates the vector components in the x-direction, depending on both \( x \) and \( y \).
• \( -2 \mathbf{j} \) represents a constant vector component in the y-direction.
• \( x z \mathbf{k} \) shows variation in the z-direction depending on \( x \) and \( z \).

Understanding vector fields is crucial for analyzing physical phenomena, as they provide insight into how a vector quantity varies across a given space.

double integrals

Double integrals are an extension of single integrals into two-dimensional spaces. Just as a single integral can calculate the area under a curve, a double integral can measure volume, surface area, or accumulate values over a two-dimensional region. They use two variables of integration, usually denoted by \( x \) and \( y \) or \( x \) and \( z \). In surface integrals, like the one in this exercise, double integrals help calculate the total across a surface rather than just along a line.
The double integral set up in the exercise is \( \iint_S 2 \, dA \), where \( dA \) is the differential area element. The limits for the integral are from \(-1\) to \(2\) for \( x \) and from \(2\) to \(7\) for \( z \). This integral essentially sums up the contributions over the specified area of the surface.
By converting this into a double integral, you are summing across a grid of points in the "rectangular" space determined by these limits. It translates the local contribution at each point to a global sum over the entire region.

rectangular coordinates

Rectangular coordinates, also known as Cartesian coordinates, are a way of representing points in space by defining their position as distances along perpendicular axes. The standard three axes used are typically the x-axis, y-axis, and z-axis. In the context of the problem, the surface we are integrating over is positioned along these axes. Specifically, it lies in the plane \(y=0\), with ranges for \(x\) from \(-1\) to \(2\) and for \(z\) from \(2\) to \(7\).

• The value \(y=0\) indicates that the surface is flat and "rests" along the xz-plane.
• The x-values go from left to right (\(-1\) to \(2\)), while z-values go upwards (\(2\) to \(7\)).

Using rectangular coordinates helps in defining surfaces precisely and makes it easier to set up and evaluate integrals in a clear, structured manner.

dot product

The dot product is an operation that combines two vectors to produce a scalar, often representing the amount of one vector in the direction of another. For vectors \( \mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k} \) and \( \mathbf{B} = B_x \mathbf{i} + B_y \mathbf{j} + B_z \mathbf{k} \), the dot product is calculated as \( \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \).
In this exercise, the dot product between the vector field \( \mathbf{F} \) and the surface's differential area vector \( d\mathbf{S} \) is computed. The surface normal is in the y-direction, specifically \( -\mathbf{j} \), making the area vector \( -dx \ dz \ \mathbf{j} \).

• When performing the dot product, the \( \mathbf{i} \) and \( \mathbf{k} \) components of \( \mathbf{F} \) disappear because they are orthogonal to \( -\mathbf{j} \).
• Only the \( -2 \mathbf{j} \) part contributes, resulting in \( 2dx \ dz \) as the dot product.

This simplified expression allows you to directly integrate over the surface to find the total flux.