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Chapter 15: Problem 1

In Exercises \(1-8,\) integrate the given function over the given surface. \( G(x, y, z)=x,\) over the parabolic cylinder \(y=x^{2}, 0 \leq x \leq 2,0 \leq z \leq 3\)

Short Answer

Expert verified
The surface integral evaluates to approximately 17.425.

Step by step solution

01

Identify the Surface

The problem describes a surface given by the equation of a parabolic cylinder, which is defined as \( y = x^2 \) within the ranges \( 0 \leq x \leq 2 \) and \( 0 \leq z \leq 3 \). This surface, therefore, is bounded in two dimensions with vertical extrusion in the z-direction.
02

Set up the Surface Integral

The surface integral over a region can typically be expressed in terms of a double integral when the surface is defined parametrically or in one of its component forms. Here, the projection of this surface on the \(xy\)-plane is a parabola between \( x = 0 \) and \( x = 2 \), and \(y\) ranges from \(0\) to \(x^2\).
03

Determine the Surface Area Element

Since the surface \( y = x^2 \) is given as a function of \(x\) and in terms of \(z\), you need \(dS = \sqrt{1 + (\frac{dy}{dx})^2} \, dx \, dz\). Here, \( \frac{dy}{dx} = 2x\), so \( dS = \sqrt{1 + 4x^2} \, dx \, dz\).
04

Set up the Double Integral

The integral of function \( G(x, y, z) = x \) over the surface becomes: \[ \int_{z=0}^{3} \int_{x=0}^{2} x \sqrt{1 + 4x^2} \, dx \, dz \].
05

Evaluate the Inner Integral

The inner integral with respect to \(x\) is \( \int_{0}^{2} x \sqrt{1 + 4x^2} \, dx \). This can often be solved by a substitution, such as setting \( u = 1 + 4x^2 \), leading to \( du = 8x \, dx \).
06

Transformation and Simplification

Using \( u = 1 + 4x^2 \), simplify the integral: the limits of \(x\) go from 0 to 2 transforming \(u\) from 1 to 17. \( dx = \frac{du}{8x} \), hence \( x dx = \frac{du}{8}\). The inner integral becomes: \[ \int_{1}^{17} \frac{\sqrt{u}}{8} \, du \].
07

Evaluate Simplified Integral

Integrate \( \int \frac{\sqrt{u}}{8} \, du = \frac{1}{8}\cdot \frac{2}{3} u^{3/2} \) evaluated from 1 to 17, which simplifies to this finite numerical evaluation.
08

Compute the Definite Integral

Substituting back, we find:\[ \frac{1}{12} [ u^{3/2} ]_{1}^{17} = \frac{1}{12} [ 17^{3/2} - 1^{3/2} ] = \frac{1}{12} [ 70.7 - 1 ] \].
09

Evaluate the Outer Integral

The factor from the solved inner integral is constant with respect to \(z\). The outer integral is simply a stretch of constant value over the \(z\) range: \[ 3 \cdot \frac{69.7}{12} \].
10

Final Calculation

Evaluate \(3 \times \frac{69.7}{12}\) for a final result. The calculation results in a final answer for the surface integral over the described region:

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Key Concepts

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

Understanding the Parabolic Cylinder
A parabolic cylinder is a fascinating geometric shape defined by a quadratic equation. In mathematics, a quadratic equation can take the form of a parabola, much like the standard form of a quadratic equation, which is often expressed as \( y = x^2 \).
  • The parabolic cylinder given in the exercise is defined by the equation \( y = x^2 \).
  • This equation describes a simple parabola in the \(xy\)-plane.
Parabolic cylinders extend infinitely in one direction, which in this exercise is the z-direction. The bounds for this particular shape are given as \( 0 \leq x \leq 2 \) and \( 0 \leq z \leq 3 \), forming a box-like shape in space with a curved side. This structure helps us conceptualize a surface in not just two dimensions but extending through the third, a key element when dealing with surface integrals in calculus.
Exploring Multivariable Calculus
Multivariable calculus builds upon single-variable calculus, adding complexity by introducing functions of two or more variables. This branch of mathematics allows us to tackle problems where variables are interdependent rather than isolated.
  • In multivariable calculus, surface integrals are crucial for integrating functions over surfaces defined in three-dimensional space.
  • The given function, \( G(x, y, z) = x \), requires integration over a surface defined parametrically by another function, \( y = x^2 \).
The challenge with multivariable calculus is managing these additional layers without losing the fundamental principles. Surface integrals rely heavily on understanding how different variables coordinate with each other to define the entire region of integration. These mathematical computations become more intuitive when visualized as composite shapes like our parabolic cylinder.
The Process of Calculus Integration
Calculus integration, especially in the context of surface integrals, involves evaluating a function over a prescribed surface. This requires both setting up the correct integral and understanding the nature of the surface in question.
  • The integral for the exercise is represented as a double integral: \[ \int_{z=0}^{3} \int_{x=0}^{2} x \sqrt{1 + 4x^2} \, dx \, dz\]
  • Surface area elements \(dS\) are derived from the derivative of the defining equation \( y = x^2 \), leading to \( dS = \sqrt{1 + 4x^2} \, dx \, dz \).
  • Integration begins with the inner integral, considering the modified limits and substituting appropriate variables for simplicity.
Through the process of substitution, and solving the integrals step-by-step, we transform complex surface area computations into manageable calculations. By understanding the integral's structure and limits, we safely navigate the arithmetic to achieve the final solution.

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Most popular questions from this chapter

Bendixson's criterion The streamlines of a planar fluid flow are the smooth curves traced by the fluid's individual particles. The vectors \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) of the flow's velocity field are the tangent vectors of the streamlines. Show that if the flow takes place over a simply connected region \(R\) (no holes or missing points \()\) and that if \(M_{x}+N_{y} \neq 0\) throughout \(R,\) then none of the streamlines in \(R\) is closed. In other words, no particle of fluid ever has a closed trajectory in \(R\). The criterion \(M_{x}+N_{y} \neq 0\) is called Bendixson's criterion for the nonexistence of closed trajectories.

In Exercises \(19-28,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\mathbf{F}=4 x \mathbf{i}+4 y \mathbf{j}+2 \mathbf{k}\) through the surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=1\) in the direction away from the z-axis

Show that if \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k},\) then \(\nabla \times \mathbf{F}=\mathbf{0}.\)

Find the flux of the ficld \(\mathbf{F}(x, y, z)=4 x \mathbf{i}+4 y \mathbf{j}+2 \mathbf{k}\) through the surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=1\) in the direction away from the \(z\) -axis.

a. Show that the outward flux of the position vector field \(\mathbf{F}=\) \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) through a smooth closed surface \(S\) is three times the volume of the region enclosed by the surface. b. Let \(n\) be the outward unit normal vector field on \(S\). Show that it is not possible for \(\mathbf{F}\) to be orthogonal to \(\mathbf{n}\) at every point of \(S .\)
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