Question

(1 point) Suppose F(x, y, z) = (x, y, 4z). Let W be the solid bounded by the paraboloid z = x2 + y2 and the plane z = 4. Let S be the closed boundary of W oriented outward. (a) Use the divergence theorem to find the flux of F through S. ſ FdA = 48pi S (b) Find the flux of F out the bottom of S (the truncated paraboloid) and the top of S (the disk). Flux out the bottom = 400pi Flux out the top =

Answer 1

TL <n, Y, 4Z) divCF) () + (9) + ? (42) 1 + 1 + 4 use cylindrical coordinates mith - - escoso, y = 8 sino such that ut ² - 2 ² OE8E2, Oco=20 Because 2= x² + y² = 4 2) x² + y² = (23² y Cirele Centered at cool with radny 2. Then according to divergenue theorem: SS Ed SSS div (F). du V 4 (6) rdz do do 0 O 22 2 =6 (012) 58 (212) do 0

1 2 2 sex [ 3) do [40 ) -( 41)] 12a [ 8-43 12 48 T Flax from top flux from cir Ciride of Z Eadius y at z=4 77 JOPI As the area Vector at top is parallel to 2-asis, so unit is :- flun through this SS (2,4,4 z). 20, 0, 1) dA R where R is region cercle utg = 4 bound by

R where Is <n,y, 48²><0,0, 17 dA R= {(8,0) / os852, orosan for a 4² rdodo } 11 32 24 (1) 2014)

So, flux out of bottom part = (48 - 32)pi = 16pi

And flux from top part = 32pi.

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