Compute the flux of \vec F (x, \ y, \ z) = 2y \hat i + 2xz \hat j through the cylindrical...


Compute the flux of {eq}\vec F (x, \ y, \ z) = 2y \hat i + 2xz \hat j {/eq} through the cylindrical surface {eq}\mathcal{S}= \{(x,y,z) \ | \ \sqrt{x^2+y^2}=4, \ x\geq 0, \ y\geq 0, \ 0\leq z \leq 4 \} {/eq} centered on the {eq}z {/eq} - axis, oriented away from the {eq}z {/eq} - axis.

Flux Integrals:

Let {eq}S {/eq} be an oriented surface in {eq}\mathbb{R}^3 {/eq} parameterized by the differentiable function {eq}s(u,v) {/eq} over some region {eq}A {/eq} in the {eq}uv {/eq}-plane. Suppose that {eq}\vec{F} {/eq} is a vector field whose domain contains all of {eq}S {/eq}. Then the flux integral of {eq}\vec{F} {/eq} through {eq}S {/eq} is defined to be

{eq}\displaystyle \iint_S \vec{F} \cdot \vec{dS}= \iint_R \vec{F}(s(u,v)) \cdot (s_u \times s_v) \, dA \, . {/eq}

If {eq}\vec{F} {/eq} represents the flow of a fluid, the flux integral of {eq}\vec{F} {/eq} through {eq}S {/eq} gives the net rate at which the fluid is passing through {eq}S {/eq} from the negative side to the positive side.

Answer and Explanation:

The surface {eq}S {/eq} is a cylinder of radius {eq}4 {/eq} centered on the {eq}z {/eq}-axis, between {eq}z=0 {/eq} and {eq}z=4 {/eq}. So we can parameterize {eq}S {/eq} in cylindrical coordinates by the function {eq}s(\theta, z)=(4 \cos \theta, 4 \sin \theta, z) {/eq} with {eq}0 \le z \le 4 {/eq} and {eq}0 \le \theta \le 2\pi {/eq}.

Differentiating {eq}r {/eq} gives:

{eq}\begin{align*} s_\theta&=-4\sin \theta \mathbf{i}+4\cos \theta \mathbf{j} \\ s_z&=\mathbf{k}\, . \end{align*} {/eq}

Computing the cross product of these two vectors then gives:

{eq}\begin{align*} s_\theta \times s_z&=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -4\sin \theta & 4\cos \theta & 0\\ 0 & 0 & 1\\ \end{vmatrix}&&\text{(by the determinant mnemonic for the cross product)}\\ &=\left[4\cos \theta(1)-0(0)\right]\mathbf{i}+\left[0(0)-(-4\sin \theta)(1)\right]\mathbf{j}+\left[-4\sin\theta(0)-4\cos \theta(0)\right]&&\text{(expanding the determinant)}\\ &=4 \cos \theta \mathbf{i}+4\sin \theta \mathbf{j} \, . \end{align*} {/eq}

Since this vector points outward from the {eq}z {/eq}-axis, the parameterization we've chosen corresponds to the correct orientation of {eq}S {/eq}.

We can therefore compute the flux of {eq}\vec{F} {/eq} as follows:

{eq}\begin{align*} \iint_S \vec{F} \cdot \vec{dS} &= \int_{z=0}^4 \int_{\theta=0}^{2\pi}\vec{F}(s(\theta, z))\cdot (s_\theta \times s_z) \, d\theta \, dz&&\text{(by the definition of the flux integral)}\\ &=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \vec{F}(4\cos \theta, 4 \sin \theta, z)\cdot (4 \cos \theta \mathbf{i}+4\sin \theta \mathbf{j}) \, d\theta \, dz&&\text{(by the above expressions for }s\text{ and }s_\theta \times s_z\text{)}\\ &=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \left[2(4\sin \theta)\mathbf{i}+2(4\cos \theta)(z)\mathbf{j}\right]\cdot (4 \cos \theta \mathbf{i}+4\sin \theta \mathbf{j}) \, d\theta \, dz&&\text{(by the given expression for }\vec{F}\text{)}\\ &=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \left(8 \sin \theta \mathbf{i}+8z\cos \theta \mathbf{j}\right)\cdot (4 \cos \theta \mathbf{i}+4\sin \theta \mathbf{j}) \, d\theta \, dz\\ &=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \left[8 \sin \theta(4\cos \theta)+8z\cos \theta(4 \sin \theta)\right] \, d\theta \, dz&&\text{(evaluating the dot product)}\\ &=\int_{z=0}^4 \int_{\theta=0}^{2\pi} 32(1+z)\sin \theta \cos \theta\, d\theta \, dz\\ &=\int_{z=0}^4\int_{\theta=0}^{2\pi} 16(1+z)\sin(2\theta) \, d\theta \, dz&&\text{(by the double-angle identity for the sine)}\\ &=\int_{z=0}^4 \Big(-8(1+z)\cos(2\theta)\Big|_{\theta=0}^{2\pi} \, dz&&\text{(evaluating the }\theta\text{-integral)}\\ &=\int_{z=0}^4 0 \, dz&&\text{(because the antiderivative in }\theta\text{ is }2\pi\text{-periodic)}\\ &=0 \, . \end{align*} {/eq}

So the flux of {eq}\vec{F} {/eq} through the given cylindrical surface is {eq}\boxed{0}\, {/eq}.

Learn more about this topic:

Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4

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