Problem 30
Question
Find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. \(\begin{array}{l}{\mathbf{F}(x, y, z)=y x^{2} \mathbf{i}-2 \mathbf{j}+x z \mathbf{k}} \\ {\text {S: rectangular surface } y=0, \quad-1 \leq x \leq 2, \quad 2 \leq z \leq 7}, \\ {\text { direction }-\mathbf{j}}\end{array}\)
Step-by-Step Solution
Verified Answer
The surface integral is 30.
1Step 1: Define the Surface and Normal Vector
The surface is given as a rectangle in the plane \( y = 0 \) with bounds \(-1 \leq x \leq 2\) and \( 2 \leq z \leq 7\). The direction specified is \(-\mathbf{j}\), indicating the negative y-direction. Therefore, the normal vector to the surface is \( \mathbf{n} = -\mathbf{j} = (0, -1, 0) \).
2Step 2: Define the Field and Parametrize the Surface
The vector field is \( \mathbf{F}(x, y, z) = yx^2 \mathbf{i} - 2 \mathbf{j} + xz \mathbf{k} \). Since the surface is already flat at \( y = 0 \), we use parameters \( x \) and \( z \) directly, with position vector \( \mathbf{r}(x, z) = (x, 0, z) \).
3Step 3: Calculate the Dot Product
Substitute \( y = 0 \) into the field \( \mathbf{F} \) to get the vector field on the surface: \( \mathbf{F}(x, 0, z) = -2 \mathbf{j} + xz \mathbf{k} \). Compute the dot product of \( \mathbf{F}(x, 0, z) \) with \( \mathbf{n} = (0, -1, 0) \): \( \mathbf{F} \cdot \mathbf{n} = 2 \).
4Step 4: Set Up the Surface Integral
The surface integral of \( \mathbf{F} \) over the surface \( S \) is given by \( \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \). Here, \( \mathbf{F} \cdot \mathbf{n} = 2 \), and the bounds for the integral are \(-1 \leq x \leq 2\) and \( 2 \leq z \leq 7\).
5Step 5: Evaluate the Integral
Calculate the integral: \[ \iint_S 2 \, dS = 2 \int_{-1}^{2} \int_{2}^{7} \, dz \, dx \]. Evaluate the inner integral: \( \int_{2}^{7} \, dz = 5 \). Now, evaluate The outer integral: \( 2 \cdot 5 \cdot \int_{-1}^{2} \, dx = 10 \cdot 3 = 30 \).
6Step 6: Combine the Results
The value of the surface integral is \( 30 \), which represents the total flux of the field \( \mathbf{F} \) over the surface in the negative y-direction.
Key Concepts
Vector FieldNormal VectorDot ProductParametrization
Vector Field
A vector field is a mathematical function that assigns a vector to every point in space. This means that for each position in the field, there is a corresponding vector that has both a direction and a magnitude. In this exercise, the given vector field is \( \mathbf{F}(x, y, z) = yx^2 \mathbf{i} - 2 \mathbf{j} + xz \mathbf{k} \).
Here, the vectors are defined in three-dimensional space using the standard unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), representing the x, y, and z axes respectively. The components of the vector field depend on the coordinates \( x \), \( y \), and \( z \).
Understanding how the vector field behaves at specific points is crucial in calculating surface integrals, as it impacts the flux through the surface. Each point on the surface feels a vector, which influences the result of the integral by determining how much field passes through the surface.
Here, the vectors are defined in three-dimensional space using the standard unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), representing the x, y, and z axes respectively. The components of the vector field depend on the coordinates \( x \), \( y \), and \( z \).
Understanding how the vector field behaves at specific points is crucial in calculating surface integrals, as it impacts the flux through the surface. Each point on the surface feels a vector, which influences the result of the integral by determining how much field passes through the surface.
Normal Vector
The normal vector is a vector that is perpendicular to a given surface at a point. In the context of surface integrals, the normal vector is important because it influences how much of the vector field flows through the surface.
For the rectangular surface defined by \( y = 0 \) in this problem, the normal vector is clearly in the negative y-direction, which is \( \mathbf{n} = -\mathbf{j} = (0, -1, 0) \). This direction is given as part of the problem statement.
The normal vector helps calculate the dot product, which is essential in determining the flow through the surface. Since the normal vector is fixed for a flat surface like this rectangle, it simplifies the calculation of the surface integral by providing a straightforward component for interaction with the vector field.
For the rectangular surface defined by \( y = 0 \) in this problem, the normal vector is clearly in the negative y-direction, which is \( \mathbf{n} = -\mathbf{j} = (0, -1, 0) \). This direction is given as part of the problem statement.
The normal vector helps calculate the dot product, which is essential in determining the flow through the surface. Since the normal vector is fixed for a flat surface like this rectangle, it simplifies the calculation of the surface integral by providing a straightforward component for interaction with the vector field.
Dot Product
The dot product is an important operation in vector analysis which allows us to multiply two vectors, resulting in a scalar. It's also referred to as the scalar product and is denoted by the symbol \( \cdot \).
In the context of surface integrals, the dot product is used to combine the vector field and the normal vector. Specifically, in this problem, you compute \( \mathbf{F}(x, 0, z) \cdot \mathbf{n} \). After substituting \( y = 0 \) into the vector field \( \mathbf{F} \), we get \( \mathbf{F}(x, 0, z) = -2\mathbf{j} + xz \mathbf{k} \).
Then, multiplying this by the normal vector \( \mathbf{n} = (0, -1, 0) \) yields the dot product \( \mathbf{F} \cdot \mathbf{n} = 2 \). This calculation essentially measures the component of the vector field that passes perpendicularly through the surface, enabling the computation of the surface integral.
In the context of surface integrals, the dot product is used to combine the vector field and the normal vector. Specifically, in this problem, you compute \( \mathbf{F}(x, 0, z) \cdot \mathbf{n} \). After substituting \( y = 0 \) into the vector field \( \mathbf{F} \), we get \( \mathbf{F}(x, 0, z) = -2\mathbf{j} + xz \mathbf{k} \).
Then, multiplying this by the normal vector \( \mathbf{n} = (0, -1, 0) \) yields the dot product \( \mathbf{F} \cdot \mathbf{n} = 2 \). This calculation essentially measures the component of the vector field that passes perpendicularly through the surface, enabling the computation of the surface integral.
Parametrization
Parametrization is a mathematical technique used to express a surface in terms of two parameters, often denoted as \( u \) and \( v \). For flat surfaces, however, the parameters can be existing coordinates.
In this exercise, since the rectangular region is already flat on \( y = 0 \), there is no need for additional transformation or conversion. We simply use \( x \) and \( z \) as parameters. Thus, the position vector for the surface becomes \( \mathbf{r}(x, z) = (x, 0, z) \).
This direct application of coordinates as parameters allows for straightforward calculation of the integral over the defined bounds from \( x = -1 \) to \( x = 2 \) and from \( z = 2 \) to \( z = 7 \). This step greatly simplifies the process, making it easier to manage the evaluation of the surface integral.
In this exercise, since the rectangular region is already flat on \( y = 0 \), there is no need for additional transformation or conversion. We simply use \( x \) and \( z \) as parameters. Thus, the position vector for the surface becomes \( \mathbf{r}(x, z) = (x, 0, z) \).
This direct application of coordinates as parameters allows for straightforward calculation of the integral over the defined bounds from \( x = -1 \) to \( x = 2 \) and from \( z = 2 \) to \( z = 7 \). This step greatly simplifies the process, making it easier to manage the evaluation of the surface integral.
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