Respuesta :
Answer:
Result;
[tex]\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = 32\pi[/tex]
Step-by-step 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
[tex]N_2 = \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} }[/tex]
Solving we have;
[tex]\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}[/tex]
Plugging the values for N₁ and N₂, we have
[tex]= \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}}{\sqrt{4x^2+4y^2+1} } d {S}[/tex]
Where:
F(x, y, z) = 2(xi +yj +zk) we have
[tex]= -\int\limits\int\limits_{S1} 2z \ dA + \int\limits\int\limits_{S2} 4x^2+4y^2+2z \ dA[/tex]
[tex]= -\int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 2z \ dA + \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2+2z \ dA[/tex]
[tex]= \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2 \ dxdy[/tex]
[tex]= \int\limits^2_{-2} \frac{(16y^2 +32)\sqrt{-(y^2-4)} }{3} dy[/tex]
= 32π.
