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Chapter 9: Problem 11

\(\operatorname{div} \mathbf{F}=2 z+10 y-2 z=10 y\) $$\begin{array}{l}\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=\iiint_{D} 10 y d V=\int_{0}^{2} \int_{0}^{2-x^{2} / 2} \int_{z}^{4-z} 10 y d y d z d x=\left.\int_{0}^{2} \int_{0}^{2-x^{2} / 2} 5 y^{2}\right|_{z} ^{4-z} d z d x \\\ \quad=\int_{0}^{2} \int_{0}^{2-x^{2} / 2}(80-40 z) d z d x=\left.\int_{0}^{2}\left(80 z-20 z^{2}\right)\right|_{0} ^{2-x^{2} / 2} d x=\int_{0}^{2}\left(80-5 x^{4}\right) d x=\left.\left(80 x-x^{5}\right)\right|_{0} ^{2}=128\end{array}$$

Short Answer

Expert verified
The surface integral evaluates to 128.

Step by step solution

01

Understanding the Problem

The problem asks us to evaluate the surface integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \) using the divergence theorem. According to the divergence theorem, this surface integral equals the volume integral \( \iiint_{D} \operatorname{div} \mathbf{F} \, dV \), where \( \operatorname{div} \mathbf{F} \) is provided as \( 10y \). Therefore, we will compute the volume integral over the region \( D \).
02

Setting Up the Volume Integral

We need to evaluate the integral \( \iiint_{D} 10y \, dV \) over the volume determined by the limits: \( x \) ranges from 0 to 2, \( z \) ranges from 0 to \( 2-x^2/2 \), and \( y \) ranges from \( z \) to \( 4-z \). The integration setup is: \( \int_{0}^{2} \int_{0}^{2-x^2/2} \int_{z}^{4-z} 10y \, dy \, dz \, dx \).
03

Integrating with Respect to y

Evaluate the innermost integral with respect to \( y \): \( \int_{z}^{4-z} 10y \, dy \). This results in \( \left. 5y^2 \right|_z^{4-z} = 5(4-z)^2 - 5z^2 = 80 - 40z \).
04

Integrating with Respect to z

Substitute the result from the \( y \)-integration into the integral with respect to \( z \): \( \int_{0}^{2-x^2/2}(80-40z) \, dz \). Evaluate this integral: \( \left. (80z - 20z^2) \right|_{0}^{2-x^2/2} = 80(2-x^2/2) - 20(2-x^2/2)^2 \).
05

Integrating with Respect to x

Evaluate the final integral with respect to \( x \): \( \int_{0}^{2}(80 - 5x^4) \, dx \). This results in \( \left. (80x - x^5) \right|_{0}^{2} = 160 - 32 = 128 \).
06

Conclusion

The value of the surface integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS = 128 \) after evaluating all the necessary steps of the integral using the divergence theorem.

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Key Concepts

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

Surface Integral
In mathematics, a surface integral allows for calculation over a surface in three-dimensional space. When dealing with vector fields, it's often represented as \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \). Here, \( \mathbf{F} \) denotes a vector field, and \( \mathbf{n} \) is the unit normal vector to the surface \( S \). The dot product \( \mathbf{F} \cdot \mathbf{n} \) signifies the flow of \( \mathbf{F} \) across the surface.
This type of integral evaluates how much of a vector field extends through a surface.
  • To compute this, project the vector field onto the normal vector of the surface.
  • Integrate this projection over the surface.
The surface integral has significant applications in physics and engineering. It's used, for instance, in calculating electric flux in electromagnetism. Utilizing the Divergence Theorem, we can transform the complex surface integral into a simpler volume integral, as demonstrated in this problem.
Volume Integral
A volume integral extends the concept of integration over three-dimensional space. It is commonly written as \( \iiint_{D} f(x, y, z) \, dV \), where \( D \) is the region in three-dimensional space and \( f(x, y, z) \) is the function to be integrated.
  • The volume integral calculates the sum of quantities over a volumetric region.
  • In spherical or cylindrical coordinates, it can be transformed to better suit the symmetry of the problem.
  • The limits of integration for each variable (often \( x, y, z \)) need careful consideration based on the region's boundaries.
In this problem, the divergence theorem simplifies our task by allowing a swap from a surface to a volume integral. Thus, evaluating over the specified limits for \( x \), \( y \), and \( z \) systematically gives the answer, demonstrating the elegance of this mathematical concept.
Mathematical Integration
Integration is a fundamental concept in calculus, representing the accumulation of quantities. While definite integrals measure the total value across limits, indefinite integrals determine antiderivatives.

In this exercise, integration takes several steps:
  • Start by integrating with respect to \( y \), simplifying the dependence on \( z \). This gives an expression that can be used in succeeding integrations.
  • Next, address the integral with respect to \( z \). Substituting limits deftly focuses the complexity into a manageable expression.
  • Finally, conclude with integration over \( x \) to find the complete solution.
This sequence highlights the importance of setting up integrals correctly and choosing the right order, often streamlining calculations and avoiding potential algebraic pitfalls.
Multivariable Calculus
Multivariable calculus extends calculus to functions of several variables. This branch includes differentiation and integration techniques for functions that depend on more than one variable.
  • The technique of changing the order of integration and limits is vital in this branch.
  • Divergence, gradient, and curl are vector operations that help analyze vector fields, encountering such concepts in multivariable calculus regularly.
  • Applications range broadly from physics to engineering, particularly in the analysis of three-dimensional systems requiring rich mathematical modeling.

The divergence theorem used here is an essential tool in multivariable calculus. It helps relate a function over a closed surface to its behavior inside the volume it encloses. This theorem not only simplifies calculations but also deepens our understanding of how functions operate over dimensions in space.

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Most popular questions from this chapter

We are given \(\sigma=k z .\) Now \(z_{x}-\frac{x}{\sqrt{16-x^{2}-y^{2}}}, \quad z_{y}=-\frac{y}{\sqrt{16-x^{2}-y^{2}}}\) \(d S=\sqrt{1+\frac{x^{2}}{16-x^{2}-y^{2}}+\frac{y^{2}}{16-x^{2}-y^{2}}} d A=\frac{4}{\sqrt{16-x^{2}-y^{2}}} d A\) Using polar coordinates, $$\begin{aligned} Q &=\iint_{S} k z d S=k \iint_{R} \sqrt{16-x^{2}-y^{2}} \frac{4}{\sqrt{16-x^{2}-y^{2}}} d A=4 k \int_{0}^{2 \pi} \int_{0}^{3} r d r d \theta \\ &=\left.4 k \int_{0}^{2 \pi} \frac{1}{2} r^{2}\right|_{0} ^{3} d \theta=4 k \int_{0}^{2 \pi} \frac{9}{2} d \theta=36 \pi k \end{aligned}$$

\(\int_{0}^{4} \int_{\sqrt{y}}^{2} \sqrt{x^{3}+1} d x d y=\int_{0}^{2} \int_{0}^{x^{2}} \sqrt{x^{3}+1} d y d x=\left.\int_{0}^{2} y \sqrt{x^{3}+1}\right|_{0} ^{x^{2}} d x\) \(=\int_{0}^{2} x^{2} \sqrt{x^{3}+1} d x=\left.\frac{2}{9}\left(x^{3}+1\right)^{3 / 2}\right|_{0} ^{2}=\frac{2}{9}\left(9^{3 / 2}-1^{3 / 2}\right)=\frac{52}{9}\)

\(R 1: y=x^{2} \Longrightarrow v+u=v-u \Rightarrow u=0\) \(\begin{aligned} R 2: y=4-x^{2} & \Longrightarrow v+u=4-(v-u) \\ & \Longrightarrow v+u=4-v+u \Longrightarrow v=2 \end{aligned}\) \(R 3: x=1 \Longrightarrow v-u=1 \Rightarrow v=1+u\) \(\frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{cc}-\frac{1}{2 \sqrt{v-u}} & \frac{1}{2 \sqrt{v-u}} \\ 1 & 1\end{array}\right|=-\frac{1}{\sqrt{v-u}}\) $$\begin{aligned} \iint_{R} \frac{x}{y+x^{2}} d A &=\iint_{S} \frac{\sqrt{v-u}}{2 v}\left|-\frac{1}{\sqrt{v-u}}\right| d A^{\prime}=\frac{1}{2} \int_{0}^{1} \int_{1+u}^{2} \frac{1}{v} d v d u=\frac{1}{2} \int_{0}^{1}[\ln 2-\ln (1+u)] d u \\ &=\frac{1}{2} \ln 2-\left.\frac{1}{2}[(1+u) \ln (1+u)-(1+u)]\right|_{0} ^{1}=\frac{1}{2} \ln 2-\frac{1}{2}[2 \ln 2-2-(0-1)]=\frac{1}{2}-\frac{1}{2} \ln 2 \end{aligned}$$

Using Green's Theorem, $$\begin{aligned} W &=\oint_{C} \mathbf{F} \cdot d \mathbf{r}=\oint_{C}-y d x+x d y=\iint_{R} 2 d A=2 \int_{0}^{2 \pi} \int_{0}^{1+\cos \theta} r d r d \theta \\ &=\left.2 \int_{0}^{2 \pi}\left(\frac{1}{2} r^{2}\right)\right|_{0} ^{1+\cos \theta} d \theta=\int_{0}^{2 \pi}\left(1+2 \cos \theta+\cos ^{2} \theta\right) d \theta \\ &=\left.\left(\theta+2 \sin \theta+\frac{1}{2} \theta+\frac{1}{4} \sin 2 \theta\right)\right|_{0} ^{2 \pi}=3 \pi \end{aligned}$$

We are given density \(=k / \rho\). $$\begin{aligned} m &=\int_{0}^{2 \pi} \int_{0}^{\infty} \int_{4 \sec \phi}^{5} \frac{k}{\rho} \rho^{2} \sin \phi d \rho d \phi d \theta=\left.k \int_{0}^{2 \pi} \int_{0}^{\cos ^{-1} 4 / 5} \frac{1}{2} \rho^{2} \sin \phi\right|_{4 \sec \phi} ^{5} d \phi d \theta \\ &=\frac{1}{2} k \int_{0}^{2 \pi} \int_{0}^{\cos ^{-1} 4 / 5}(25 \sin \phi-16 \tan \phi \sec \phi) d \phi d \theta \end{aligned}$$ $$\begin{aligned} &=\left.\frac{1}{2} k \int_{0}^{2 \pi}(-25 \cos \phi-16 \sec \phi)\right|_{0} ^{\cos ^{-1} 4 / 5} d \theta=\frac{1}{2} k \int_{0}^{2 \pi}[-25(4 / 5)-16(5 / 4)-(-25-16)] d \theta\\\ &=\frac{1}{2} k \int_{0}^{2 \pi} d \theta=k \pi \end{aligned}$$
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