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In Exercises \(35-44,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the given direction. Parabolic cylinder \(\mathbf{F}=x^{2} \mathbf{j}-x z \mathbf{k}\) outward (normal away from the \(y z\) -plane) through the surface cut from the parabolic cylinder \(y=x^{2},-1 \leq x \leq 1,\) by the planes \(z=0\) and \(z=2\)

Short Answer

Expert verified
The flux is \( \frac{4}{3} \).

Step by step solution

01

Understand the Problem

We need to find the flux of the vector field \( \mathbf{F} = x^2 \mathbf{j} - xz \mathbf{k} \) across a parabolic cylinder defined by \( y = x^2 \) for \( -1 \leq x \leq 1 \), bounded by the planes \( z = 0 \) and \( z = 2 \). The normal direction is outward, away from the \( yz \)-plane.
02

Parametrize the Surface

We will use a parametrization \( \mathbf{r}(x,z) = \langle x, x^2, z \rangle \) for portions of the surface on the cylinder. Since the surface is cut by \( z = 0 \) and \( z = 2 \), our parameters are \( x \) in \( [-1, 1] \) and \( z \) in \( [0, 2] \).
03

Find the Normal Vector

Calculate the tangent vectors to the surface: \( \frac{\partial \mathbf{r}}{\partial x} = \langle 1, 2x, 0 \rangle \) and \( \frac{\partial \mathbf{r}}{\partial z} = \langle 0, 0, 1 \rangle \). The normal vector \( \mathbf{n} \) is given by their cross product: \( \mathbf{n} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial z} = \langle -2x, 1, 0 \rangle \). Since the outward normal is away from the \( yz \)-plane, this normal is correct.
04

Compute the Dot Product \( \mathbf{F} \cdot \mathbf{n} \)

Substitute \( \mathbf{F} = x^2 \mathbf{j} - xz \mathbf{k} \) and \( \mathbf{n} = \langle -2x, 1, 0 \rangle \). The dot product \( \mathbf{F} \cdot \mathbf{n} \) becomes \( x^2 \cdot 1 - xz \cdot 0 = x^2 \).
05

Set Up and Evaluate the Double Integral

The flux integral is given by \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma = \int_{-1}^{1} \int_{0}^{2} x^2 \, dz \, dx \). Compute the integral with respect to \( z \) first, which gives \( \int_{0}^{2} x^2 \, dz = 2x^2 \). Substitute this into \( \int_{-1}^{1} 2x^2 \, dx \), and evaluate: \( \int_{-1}^{1} 2x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_{-1}^{1} = 2 \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = 2 \left( \frac{2}{3} \right) = \frac{4}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametrization
Parametrization is a technique used to express a surface or curve in terms of parameters, making it easier to perform calculations such as integration. In this exercise, we are working with a surface cut from a parabolic cylinder, defined by equations relating to variables like in functions. Specifically, the surface is given by the parabolic equation \( y=x^2 \). To parametrize it, we select two parameters, \( x \) and \( z \). Here, \( x \) ranges from \(-1\) to \(1\), and \( z \) ranges from \(0\) to \(2\). These ranges are derived from the constraints and boundaries given in the problem. So, the surface can be described by the vector function \( \mathbf{r}(x,z) = \langle x, x^2, z \rangle \).

This function traces every point on the surface as the parameters vary within their specified ranges. In simple terms, parametrization transforms our surface into a version we can use to integrate the dot product of the given vector field with a normal vector, which gives us the flux.

By breaking down the surface into simpler, manageable portions, parametrization is a powerful tool for solving otherwise complex calculus problems.
Parabolic Cylinder
A parabolic cylinder is a three-dimensional surface that is extended along one axis, forming a cylinder with a parabolic cross-section. The specific cylinder in this exercise is defined by the equation \( y = x^2 \). This means that for each value of \( x \), the corresponding \( y \) value forms a parabola, and these parabolas are repeated along the \( z \)-axis, forming a "cylinder" of sorts.

Here are a few key insights into parabolic cylinders:
  • They are formed by extruding a parabola usually along the third dimension (in this case, the \( z \)-direction).
  • The bounded area on which we operate is only between \( z=0 \) and \( z=2 \). It confines the open surface to a measurable segment, essential for calculating flux.
When visualizing this, imagine slicing through the surface vertically (with constant \( x \))—you'll see a parabolic curve, but extend these curves along the \( z \) direction to form a cylindrical shape.

This transformation in three-dimensional space defines the region over which we seek to calculate the flux of our vector field.
Vector Fields
Vector fields are mathematical constructs where a vector is assigned to every point in a given space. In our exercise, the vector field is expressed as \( \mathbf{F} = x^2 \mathbf{j} - xz \mathbf{k} \), where each point in space has direction and magnitude defined by this equation. Here are critical aspects of using vector fields in flux calculations:

  • Vector fields have components that can be functions of coordinates, e.g., \( x^2 \) and \(-xz\) in this case.
  • Understanding the direction of these vectors at every point is crucial for determining flux, as they interact with surfaces.
To calculate flux, we need the interaction of this field with a surface, which is characterized by the dot product with the normal vector of the surface. In simpler terms, flux represents how much of the vector field "flows" through the surface, factoring in direction and strength.

By computing this, we account for only the component of the vector field that goes through the surface, giving us a measure that reflects the distribution of vectors across the area of interest. Hence, understanding how vector fields operate and their representation is essential when working with them in the context of flux calculations and other applications.

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Most popular questions from this chapter

Find the area of the surface \(x^{2}-2 \ln x+\sqrt{15} y-z=0\) above the square \(R : 1 \leq x \leq 2,0 \leq y \leq 1,\) in the \(x y\) -plane.

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In Exercises \(41-44\) , use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(\mathbf{F}\) around the simple closed curve \(C .\) Perform the following CAS steps. \(\mathbf{F}=x e^{y} \mathbf{i}+4 x^{2} \ln y \mathbf{j}\) \(C :\) The triangle with vertices \((0,0),(2,0),\) and \((0,4)\)

Green's first formula Suppose that \(f\) and \(g\) are scalar functions with continuous first- and second-order partial derivatives throughout a region \(D\) that is bounded by a closed piecewise-smooth surface \(S .\) Show that $$\iint_{S} f \nabla g \cdot \mathbf{n} d \sigma=\iiint_{D}\left(f \nabla^{2} g+\nabla f \cdot \nabla g\right) d V$$ Equation (9) is Green's first formula. (Hint: Apply the Divergence Theorem to the field \(\mathbf{F}=f \nabla g . )\)

Flux of a gradient field Let \(S\) be the surface of the portion of the solid sphere \(x^{2}+y^{2}+z^{2} \leq a^{2}\) that lies in the first octant and let \(f(x, y, z)=\ln \sqrt{x^{2}+y^{2}+z^{2}}\) . Calculate $$ \iint_{S} \nabla f \cdot \mathbf{n} d \sigma $$ \((\nabla f \cdot \mathbf{n}\) is the derivative of \(f\) in the direction of \(\mathbf{n} . )\)

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