Question

Use the Divergence Theorem to evaluate the following integral S F · N dS and find the outward flux of F through the surface of the solid bounded by the graphs of the equations. Use a computer algebra system to verify your results. F(x, y, z) = 2(x???? + y???? + z????) S: z = 0, z = 4 − x2 − y2

Answer 1

Result;

`\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = 32\pi`

Explanation:

Where:

F(x, y, z) = 2(x·i +y·j +z·k) and

S: z = 0, z = 4 -x² - y²

For the solid region between the paraboloid

z = 4 - x² - y²

div F

For S: z = 0, z = 4 -x² - y²

We have the equation of a parabola

To verify the result for F(x, y, z) = 2(x·i +y·j +z·k)

We have for the surface S₁ the outward normal is N₁ = -k and the outward normal for surface S₂ is N₂ given by

`N_2 = \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{√(4x^2+4y^2+1) }`

Solving we have;

`\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = \int\limits\int\limits_(S1) { \textbf{F}} \, \cdot \textbf{N}_1 d {S} + \int\limits\int\limits_(S2) { \textbf{F}} \, \cdot \textbf{N}_2 d {S}`

Plugging the values for N₁ and N₂, we have

`= \int\limits\int\limits_(S1) { \textbf{F}} \, \cdot \textbf{(-k)}d {S} + \int\limits\int\limits_(S2) { \textbf{F}} \, \cdot \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{√(4x^2+4y^2+1) } d {S}`

Where:

F(x, y, z) = 2(xi +yj +zk) we have

`= -\int\limits\int\limits_(S1) 2z \ dA + \int\limits\int\limits_(S2) 4x^2+4y^2+2z \ dA`

`= -\int\limits^2_(-2) \int\limits^(√(4-y^2)) _(-√(4-y^2)) 2z \ dA + \int\limits^2_(-2) \int\limits^(√(4-y^2)) _(-√(4-y^2)) 4x^2+4y^2+2z \ dA`

`= \int\limits^2_(-2) \int\limits^(√(4-y^2)) _(-√(4-y^2)) 4x^2+4y^2 \ dxdy`

`= \int\limits^2_(-2) ((16y^2 +32)√(-(y^2-4)) )/(3) dy`

= 32π.