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Additional Exercises 5

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Department
Mathematics
Course
MAT237Y1
Professor
All Professors
Semester
Fall

Description
1. (a) [6 marks] Let F(x, y,z) = y i + (2xy −e )j+3z k 2 (i) Show that F is conservative on R . 1 3 3 The vector field F is of class C on R , and of course R is simply connected. Hence F is conservative iff curlF = 0 which actually is the case i j k curlF = ∂ ∂ ∂ = (0 − 0, 0+ 0,2 y −2 )= 0 ∂x ∂y ∂z 2 2y 2 y 2xy − e 3 z (ii) Find the potential function for F i.e. a scalar function f such that ∇f . ∂f ∂f ∂ f 2 2y 2 We should have ( , , ) = (y ,2xy −e ,3z ) and ∂x ∂ y ∂ z ∂f 2 2 ∂f 2y = y implies f (x, y,z) = xy + h(y,z), substituting to = 2xy −e we get ∂x ∂y ∂h 1 2xy + = 2xy −e 2y which implies h(y,z) = − e 2y+ k(z) so ∂y 2 1 ∂f f (x, y,z) = xy − e 2y+ k(z). Substituting this to = 3z we get k (z.) = z + c 2 ∂z 1 Hence f x( , , )xy − e2 2y+ z +c 2 2 2y 2 1.(b) [6 marks] Evaluate G∫dx where G(x, y,z) = (y + y)i +(2xy −e )j+3z k and C C is a helix parametrized by g(t) = (cost,sint,t), 0 ≤ t ≤ π . HINT: Use the first part and write G = ∇f + H 2 1 2y 3 We may write G(x, y,z) = ∇(xy − e + z ) +(y,0,0). For t = 0we get the point (1,0,0) and 2 for t =π we get the point (−1,0,π) . Hence 2 1 2y 3 2 1 2y 3 ∫ G⋅dx = ∫ ∇(xy − 2e + z )⋅dx + (∫,0,0)⋅dx = ∫∇(xy − 2 e + z )⋅dx + ∫dx = C C C C C 1 π 1 1 π = [xy − e2y + z ](1,0,0) (sint)(−sint)dt = [(0 − +π 3) −0 − + 0)]− = 2 0 2 2 2 3 π = π − 2 2 2. [10 marks] Verify the Divergence Theorem in case F(x, y,z) = xi + zj and V is the region in the first octant bounded by the plane 2 x + 2 + z = 4 and the coordinate planes. The boundary S of V consists of four faces S x=0 ,S y=0, Sz=0and S plane(i.e. it is piecewise smooth). Hence we have, taking outward normals ∫∫ F⋅ndS = ∫∫F⋅ndS + ∫∫ F⋅ndS + ∫∫F⋅ndS + ∫∫F⋅ndS = ∫∫(x,z,0)⋅(−1,0,0)dS + S x=0 Sy=0 Sz=0 Splane Sx=0 + ∫∫(x,z,0)⋅(0,−1,0)dS + ∫∫(x,z,0)⋅(0,0,−1)dS + ∫∫(x,z,0)⋅ndS = Sy=0 Sz=0 Splane 2 4−x 2 2−x = 0− zdzdx + 0 + (x,4 − 2x − 2y,0)⋅(2,2,1)dydx = ∫∫ ∫∫ 0 0 0 0 1 2 2 2−x = − (4 − 2 )dx + 2 (4 − x − 2y)dydx = 2 0 0∫0 2 1 1 (4 2 ) 3 2 2 = 0 + 2∫[4(2 − x) − x(2 − x) − (2 − x) ]dx = 2 3 2 0 16 16 8 = − + 2[4x − x ]0= − +8 = 3 3 3 1 1 8 On the other hand we have (divF)dV = (1+0+0)dV = dV = B h = 2 4= . V∫∫ ∫V∫ ∫V∫ 3 3 3 If you do not use the formula for the volume, you have to evaluate the iterated integral 2 2−x4 2x−2y 8 dV = dzdydx = ∫V∫ 0∫ 0 0 3 3. (a)[2 marks] Let S b0 a smooth surface in R and let S ⊂ S . Defin0 the interior point x of S relative to S 0 3 A point x∈S is the interior point oS relative to S 0f it has a neighborhood U ⊂ R such that U ∩S ⊂ S . 0 (b) [7 marks] Evaluate (∫ +sin x)dx +(z +cos y)dy + xdz , where C
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