Let F = [x^2,0,z^2], and S the surface of the box abs(x) <= 1, abs(y) <= 3, 0 <= z <= 2. Evaluate the surface integral int int_S F*n dA by the divergence theorem.

Question

Let

F = [x^{2}, 0, z^{2}]

, and S the surface of the box

| x | \leq 1, | y | \leq 3, 0 \leq z \leq 2

.
Evaluate the surface integral

\int \int_{S} F \cdot n d A

by the divergence theorem.

Answer 1

Step 1
The divergence of the function F is:
$\nabla \cdot F = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$
$\div F = \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial y} (0) + \frac{\partial}{\partial z} (z^{2})$
$= 2 x + 2 z$
Step 2
By the divergence theorem convert the surface integral into a triple integral:
$\int \int_{S} P d y d z + Q d x d z + R d x d y = \int \int_{V} \int (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) d x d y d x$
$= \int \int_{V} \int (2 x + 2 z) d x d y d z$
Step 3
$\int \int_{V} \int (2 x + 2 z) d x d y d z = \int_{- 1}^{1} d x \int_{3}^{3} d y \int_{0}^{2} (2 x + 2 z) d z$
$= 2 \int_{1}^{1} d x \int_{3}^{3} d y \in_{0}^{2} (x + z) d z$
$= 2 \int_{1}^{1} d x \int_{3}^{3} d y [(x z + \frac{z^{2}}{2}) ∣_{0}^{2}]$
$= 2 \int_{1}^{1} d x \int_{3}^{3} (2 x + 2) d y$
$= 4 \int_{1}^{1} d x [(x y + y) ∣_{- 3}^{3}]$
$= 4 \int_{1}^{1} d x [(3 x + 3) - (- 3 x - 3)]$

This is helpful 78

Let F = [x^2,0,z^2], and S the surface of the box abs(x) <= 1, abs(y) <= 3, 0 <= z <= 2. Evaluate the surface integral int int_S F*n dA by the divergence theorem.

Want to know more about Green's, Stokes', and the divergence theorem?

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