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Chapter 15: Problem 24

Calculate the outward flux of the vector field F across the given closed surface \(S\). \(\mathbf{F}=x^{2} \mathbf{i}+2 y^{2} \mathbf{j}+3 z^{-} \mathbf{k}: \quad S\) is the boundary of the solid bounded by the cone \(z=\sqrt{x^{2}+y^{2}}\) and the plane \(z=3\)

Short Answer

Expert verified
The outward flux of \( \mathbf{F} \) across the surface \( S \) is 81\( \pi \).

Step by step solution

01

Understand the Flux Integral

To calculate the outward flux of a vector field \( \mathbf{F} \) across a closed surface \( S \), we use the surface integral \( \oint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) is the outward unit normal vector to the surface, and \( dS \) is the differential surface area.
02

Apply the Divergence Theorem

Since \( S \) is a closed surface, we can simplify the computation using the Divergence Theorem: \( \oint_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{V} abla \cdot \mathbf{F} \, dV \), where \( V \) is the volume enclosed by \( S \).
03

Calculate the Divergence of F

Compute the divergence of the vector field \( \mathbf{F}(x, y, z) = x^2 \mathbf{i} + 2y^2 \mathbf{j} + 3z \mathbf{k} \). The divergence is \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(2y^2) + \frac{\partial}{\partial z}(3z) = 2x + 4y + 3 \).
04

Set Up the Triple Integral

The volume \( V \) is bounded by the cone \( z = \sqrt{x^2 + y^2} \) and the plane \( z = 3 \). Convert the integral to cylindrical coordinates: \( x = r \cos \theta \), \( y = r \sin \theta \), \( z = z \). The limits are \( 0 \leq r \leq 3 \), \( 0 \leq \theta \leq 2\pi \), and \( r \leq z \leq 3 \).
05

Evaluatethe Triple Integral

Insert the divergence into the triple integral and solve: \( \iiint_{V} (2x + 4y + 3) \, dV = \int_{0}^{2\pi} \int_{0}^{3} \int_{r}^{3} (2r \cos \theta + 4r \sin \theta + 3) \cdot r \, dz \, dr \, d\theta \). Evaluate this integral step-by-step.

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Key Concepts

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

Divergence Theorem
The Divergence Theorem, also known as Gauss's Theorem, is a powerful tool in vector calculus that connects the flux across a closed surface to a volume integral over the region it encloses. This theorem is particularly useful when dealing with the outward flux of a vector field across a closed surface. In mathematical terms, the Divergence Theorem states:
  • If \( V \) is a region in space bounded by a smooth closed surface \( S \), then the surface integral over \( S \) of the vector field \( \mathbf{F} \) is equal to the volume integral over \( V \) of the divergence of \( \mathbf{F} \).
  • Essentially, it transforms a difficult surface integral into a generally easier volume integral.
This is particularly advantageous when the divergence \( abla \cdot \mathbf{F} \) is simple or zero, allowing us to compute the flux without directly evaluating the surface integral.
Vector Field
A vector field assigns a vector to every point in a subset of space. In this exercise, the given vector field is \( \mathbf{F}(x, y, z) = x^2 \mathbf{i} + 2y^2 \mathbf{j} + 3z \mathbf{k} \). Here, each term represents the influence along the respective axes:
  • \( x^2 \mathbf{i} \) affects the field along the x-axis.
  • \( 2y^2 \mathbf{j} \) affects the field along the y-axis.
  • \( 3z \mathbf{k} \) influences the field along the z-axis.
To understand how the field behaves, we calculate its divergence. This operation helps us understand how much the field spreads out or converges at different points, which is pivotal when relating to flux.
Cylindrical Coordinates
Cylindrical coordinates are a three-dimensional coordinate system that prescribes a point by its distance from a chosen axis (radial distance), its angle around that axis, and its height along that axis. The transformation from Cartesian to cylindrical coordinates is given by:
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
  • \( z = z \)
This system is particularly useful for regions like cylinders or cones where symmetry makes it easier to set up and compute integrals, as it reduces the complexity found in Cartesian coordinates. In the provided exercise, converting to cylindrical coordinates simplifies the limits for the triple integral, making the evaluation feasible.
Triple Integral
A triple integral allows you to take into account variations in all three spatial dimensions, which is crucial for calculating quantities such as volume, mass, and total flux in three-dimensional regions. Here, the triple integral is set to calculate the flux of the given vector field:
  • The integrand, \( 2x + 4y + 3 \), represents the divergence of \( \mathbf{F} \), a necessary component when applying the Divergence Theorem.
  • The region \( V \) over which the integration is performed is bounded by the equations of a cone and a plane, leading to precise limits of integration.
  • These bounds are transformed into cylindrical coordinates to easily describe the solid geometric region.
By evaluating the triple integral, you find the total flux through the surface \( S \), a value that informs us about how the vector field is behaving within the enclosed volume.

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Most popular questions from this chapter

Show that if \(\mathbf{F}\) is a constant force field, then it does zero work on a particle that moves once uniformly counterclockwise around the unit circle in the \(x y\) -plane.

The Laplacian of the twice-differentiable scalar function 7 is defined to be \(\boldsymbol{r}^{2} f=\operatorname{div}(\operatorname{grad} f)=\boldsymbol{\nabla} \cdot \boldsymbol{\nabla} f .\) Show that $$\boldsymbol{\nabla}^{2} f=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}$$

Evaluate the surface integral \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S\), where \(\mathbf{n}\) is the upward-pointing unit normal vector to the given surface \(S\). \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}: \quad S\) is the hemisphere \(z=\sqrt{9-x^{2}-y^{2}}\).

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Evaluate the line integral $$\int_{C} \mathbf{F} \cdot d \mathbf{r}=\int_{C} \mathbf{F} \cdot \mathbf{T} d s$$ along the indicated path \(C\). \(\mathbf{F}(x, y, z)=y z \mathbf{i}+x z \mathbf{j}+x y \mathbf{k}: \quad C\) is the straight line seg- ment from \((2,-1,3)\) to \((4,2,-1)\).
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