Question

Use the divergence theorem to calculate the surface integral $iint_S mathbf{F} cdot dmathbf{S}$; that is, calculate the flux of $mathbf{F}$ across $S$. $mathbf{F}(x, y, z) = x^2 z mathbf{i} + xz^3 mathbf{j} + y ln(x + 1) mathbf{k}$, $S$ is the surface of the solid bounded by the planes $x + 2z = 8$, $y = 3$, $x = 0$, $y = 0$, and $z = 0$

          Use the divergence theorem to calculate the surface integral $iint_S mathbf{F} cdot dmathbf{S}$; that is, calculate the flux of $mathbf{F}$ across $S$.
$mathbf{F}(x, y, z) = x^2 z mathbf{i} + xz^3 mathbf{j} + y ln(x + 1) mathbf{k}$, $S$ is the surface of the solid bounded by the planes $x + 2z = 8$, $y = 3$, $x = 0$, $y = 0$, and $z = 0$

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Use the divergence theorem to calculate the surface integral iintS mathbfF cdot dmathbfS; that is, calculate the flux of mathbfF across S.
mathbfF(x, y, z) = x^2 z mathbfi + xz^3 mathbfj + y ln(x + 1) mathbfk, S is the surface of the solid bounded by the planes x + 2z = 8, y = 3, x = 0, y = 0, and z = 0

Added by Kimberly W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Israel Hernandez

09/17/2023

Step 1

From the given planes, we have: z + 2z = 3 3z = 3 z = 1 So, the solid is bounded by the planes z = 0, y = 3, x = 0, and z = 1. In the xy-plane, this corresponds to the region R bounded by the lines y = 3, x = 0, and y = 0. Show more…

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Use the divergence theorem to calculate the surface integral; that is, calculate the flux of F(x, y, z) = x^2zi + xz^3j + y ln(x + 1)k across S. S is the surface of the solid bounded by the planes x + 2z = 8, y = 3, x = 0, y = 0, and z = 0.

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00:01 So in this problem, we are given a vector field.

00:03 Let us write this out.

00:04 The vector field f has three variables, x, y, and z.

00:09 And it is equal to x squared times z times i hat plus x times z to the power of 3 times j hat and then plus y times ln of x plus 1 times k hat.

00:28 And the divergence of this vector field is going to equal to partial partial x of x squared times z plus partial partial y of xz to the power of 3 and then plus partial partial z of y times ln of x plus 1.

00:49 And therefore, we're going to get the divergence to simply be equal to 2xz.

00:56 Notice that the last two terms are simply equal to 0.

01:01 And next, the double integral over the surface s of f dotted with ds is now going to equal to, by the divergence theorem, the triple integral over the solid e of the divergence of the vector field dv.

01:26 So for that, we want to make sure that we set this up correctly.

01:30 So we know that z is bounded below by 0 and bounded above by the plane.

01:35 The plane here is x plus 2z equals 8.

01:39 We're solving for z here.

01:41 We get that z is going to equal to negative x over 2 plus 4.

01:47 So that will be negative x over 2 plus 4.

01:51 What about y? y here ranges between the values of 0 and 3.

01:57 That is also given...

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