237t3s - 1. Let S be the portion of the cone z 2 = x 2 + y...

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1. Let S be the portion of the cone with 2 2 2 y x z + = 2 0 z and . 0 x (a) [2 marks] Parametrize the surface S using cylindrical coordinates. 3 2 : R R D G where } 2 2 , 2 0 : ) , {( ), , sin , cos ( ) , ( π θ = = r r D r r r r G (b) [6 marks] Compute dS x y x S ) ( 2 2 + + ∫∫ r r r r r r r 2 || ) , sin , cos ( || || 0 cos cos 1 sin cos || || || = = = × k j i G G and we have = dS x y x S ) ( 2 2 + + ∫∫ drd r r r D 2 ) cos ( 2 + ∫∫ = d dr r r ) cos ( 2 2 2 2 2 0 3 + ∫∫ = = d r r 2 0 3 2 2 4 ] cos 3 4 [ 2 + = d ) cos 3 8 4 ( 2 2 2 + = 2 2 ] sin 3 8 4 [ 2 + = ) 4 3 ( 3 2 4 + 2
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2. [8 marks] Evaluate ∫∫ , where F dS curl S n F ) ( k i xy yz + = and S is the part of the paraboloid between the planes 2 2 z y x + = 1 = x and 4 = x , n pointing away from the x -axis . The integral may be evaluated directly, but it is much simpler to use Stoke’s Theorem. where C is the circle with counterclockwise orientation as viewed 2 1 C C S = 1 1 2 2 = + y x from the positive x -axis, and is the circle with clockwise orientation as 2 C 4 2 2 = + y x viewed from the positive x -axis. The parametrizations of those oriented circles are : 1 C π 2 0 , ) sin , cos , 1 ( ) ( = t t t t g and C : 2 2 0 , ) cos 2 , sin 2 , 4 ( ) ( = t t t t g By Stoke’s Th. we get ∫∫ = + = + = dS curl S n F ) (= S d x F 1 C d x F 2 C d x F dt t 2 0 2 cos dt t ) sin 16 ( 2 0 2 15 3
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3. [9 marks] Find the flux of across the part of ) 1 , , ( ) , , ( 2 2 2 + = z ye xe z y x z z F the sphere with 2 2 2 2 = + + z y x 0 , 0 z y , n pointing out of the origin. HINT: Direct calculation might be a failure, close the surface. Close the given surface (which is a quarter sphere) by and being the projections of on the coordinate planes y = 0 and z = 0 respectively. Consider . 1 S 2 S 3 S 1 S 3 2 1 S S S S = Since S is piecewise smooth closed surface enclosing a quarter ball and q B ) ( 1 q B C F we can use the Divergence Theorem and then spherical coordinates to evaluate the resulting integral ∫∫ = = = dS S n F ∫∫∫ q B dV div F ∫∫∫ q B dV z 2 θ φ ρ ππ π d d d ) sin )( cos ( 2 2 0 2 2 0 ∫∫∫ = = ) )( 2 sin ( 2 2 0 3 d d ∫∫ = = 2 0 4 2 ] 4 [ ] 2 2 cos [ . Now we have that dS S n F ∫∫ 1 + + = dS S n F ∫∫ 2 dS S n F ∫∫ 3 where n is the outer normal to S , and since ∫∫ = = and = = dS S n F 2 dS S ) ( 2 j F ∫∫ 0 0 2 = ∫∫ dS S dS S n F ∫∫ 3 dS S k F ∫∫ 3 = = ∫∫ 2 ) 2 ( 2 1 1 3 dS S the flux across is equal to = 1 S dS S n F ∫∫ 1 2 = 4
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4. (a) [2 marks] Determine the largest set(s) 2 R D
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237t3s - 1. Let S be the portion of the cone z 2 = x 2 + y...

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