#10 Using the Divergence Theorem Is Exercise 7-16, the Divergence Theorem to eva

Question

#10

Using the Divergence Theorem Is Exercise 7-16, the Divergence Theorem to evaluate integral_S integral F - N dS and find the outward flux of F through the surf bounded by the graphs of the equations. Use a algebra system to verify your results. F(x, y, z) = x^2 i + y^2 j + z^2 k S: x = 0, y = a, z = 0, z = a FF(x, y, z) = x^2 z^2 i - 2y j + 3xyzk S: x = 0, x = a, y = 0, y = a, z = 0, z = a F(x, y, z) = x^2 i - 2xyj + xyz^2 k S: z = squareroot a^2 - x^2 - y^2, z = 0 F(x, y, z) = xyi + yzj - yzk S: z = squareroot a^2 - x^2 - y^2, z = 0

Explanation / Answer

F=<xy ,yz ,-yz>

divF=y +z -y

divF=z

by divergence theorem flux =divF dv

z=[a2-x2-y2] ,z=0

[a2-x2-y2]=0

x2+y2=a2

in cylindrical coordinates

x=rcos y=rsin

x2+y2=r2

0<=<=2,0<=r<=a ,0<=z<=[a2-r2]

dv =r dz dr d

flux =[0 to 2] [0 to a] [0 to [a2-r2]] z r dz dr d

flux =[0 to 2] [0 to a][0 to [a2-r2]] (1/2)z2 r dr d

flux =[0 to 2] [0 to a](1/2)[a2-r2]2 r dr d

flux =[0 to 2] [0 to a](1/2)[a2-r2] r dr d

flux =[0 to 2] [0 to a](1/2)[ra2-r3] dr d

flux =[0 to 2][0 to a](1/2)[(1/2)r2a2-(1/4)r4] d

flux =[0 to 2](1/2)[(1/2)a2a2-(1/4)a4] -0 d

flux =[0 to 2](1/8)a4 d

flux =[0 to 2](1/8)a4

flux = (1/8)a4(2-0)

flux = (/4)a4

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