Chapter 16: Problem 30
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 \(-\mathbf{j}\)
Short Answer
Step by step solution
Express Surface Parameters
Define the Surface Element
Calculate the Dot Product
Set Up the Double Integral
Solve the Integral
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
vector fields
- \( y x^{2} \mathbf{i} \) 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 \( z \).
double integrals
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
- 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 \(7\)).
dot product
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.