Question

4.8) a) Complete the statement of: The Divergence Theorem: Let D be a closed solid in space bounded by a closed surface s oriented by an outwardly directed unit normal vector n. If F(x, y, z)=(M(x,y,z), N(x, y, z), P(x, y, z)) where M, N, and P have continuous partial derivatives in D, then: D b) Use the Divergence Theorem to write as an iterated integral the flux of F=(x",x’y,x?:) over the closed cylindrical surface whose sides are defined by the equation x + y = 4, and whose top and bottom are = 0 and 2 = 3, r° +y<4. Do not evaluate the integral.

Answer 1

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Here i used divergence theorom and not evaluate integral as say in question

solution: (a); Let D be a closed solid in space Bounded By By outwardly directed a closed surface S oriented an unit Normal vector n. if where F(8,4, 2) = < M (X,Y,Z), NO3, y, z), P(x, y, z)) M, N, P have continuous partial derivatives then. SS Fonds SSS o.Ě dv. S D Answer SS (M € + Nỹ + pâlon ds SSS Comment same) av Solve! Now to use divergence theorom to write integral the flux of f =<x3, x²y ; x²2) over the Closed cylindrica). Surface whose sides are defined By the equation, re? + y2 = 4 ; Whose Bottom are 2-0, 2-3, 22+y? <4. top & first to draw the surface, Y у q?+y?= 4. 22 x →x Z-0, S z 2 2-3 (0,2) 2²+y²=4 x-y plane projection on (-2,0) (2,0) (0,-2) Now Flux SS (F)inds 2 S 0

Νοω F = <x3, xy, x?z.). ane So continuous & all x3, x²y, x²z polynomia) it is always have continuous partial derivative. Divergence theorom applicable SS Finds SSS v. Ē dv. S V where vi's Solid. By 2=0 to 2=3, 22+42 54. flux = SSS o. Ē dv. V Νοω VF = a (x3) + dx & cx?y) + ду (222) o2 3x² + x² + x²= 5x2. flux SSS 5x2 dv. VE 3 {(x,y,2) 03253, - Mu-x < g < Ju-xº, -2 < x < 2 3. (4-2² flur = s s s sx? dz dy dze. -2 -fu-x? Ń ON 2 14-7² 3 flux ss , 5x² dz dy dz Answei -2 (4-2² coordinates Now to simply fy the integral By changing the into cylindrical coordinates. let x = raso, y= rsino, z = Z. → {(7,0,2) :05233, 05152, os se o < 203 & dz dy dve z dzdrdo So we have the flux will changed integral.

2TT 2 flux = 5(0 coso)? z dz dodo E no ha O 2 3 S S S 573 cos?o dz do do 2 IT 2 3 → flux = n S S S (583 Cos?0) dz do do 8=0 2-0 Answer
in second part i provide you two answer one in x,y,z coordinate second in cylinerical coordinates