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MAT237Y1 (47)
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School
University of Toronto St. George
Department
Mathematics
Course
MAT237Y1
Professor
all
Semester
Fall

Description
1. Suppose F(x, y,z) = azi + xj+ yk where a is a constant. (a) [5 marks] Is there a real number a such that div[(curlF)×F] > 0. Justify. We have i j k i j k ∂ ∂ ∂ curlF = = (1,a,1) and (curlF)×F = 1 a 1 = (ay − x,az − y,x − a z) ∂x ∂y ∂z az x y az x y 2 Hence div [(rlF ×) =− ] − 1 1 0 for all values of a and so there is no value of a such that div[(urlF ×) > ] 0. (b) [7 marks] In case a = −2 , evaluate ∫F⋅dx , where C is the portion of the curve of C intersection of surfaces z = x2and x + y = 4 in the first octant from(2t,0,4) o (.,2,0) In this case F x y z ) = −2zi xj+ yk From x + y = 2 and z = x 2 (both equations should 2 be satisfied) we get a required parametrization of C : x = 2cost , y = 2sint, z = 4cos t , π 0 ≤ t ≤ and we have 2 π / 2 F⋅dx = (−8cos t, 2cost,2sint)⋅(−2sint, 2cost, −8costsint)dt = C 0 π / 2 2 2 2 = ∫(16cos t sint + 4cos t −16sin t cost)dt = 0 16 3 16 3 π / 2 16 16 = [− cos t+ 2t +sin2t − sin t]0 = (0+π + 0− )−(− + 0+0−0) = π 3 3 3 3 2 2. [8 marks] Use Green’s theorem (check if applicable!) to find the work done by the vector 2 field F(x, y) = (y + e )i + (3x + y)y j in moving a particle from the origin along the 2 2 upper semicircle x + y = x y ≥ 0 to the point (2,0), then along the line segment from (2,0) to(0,−2) , and then from (0,−2) back to the origin along y-axis. Consider the region D enclosed by the path followed by the particle. The region D is 1 2 regular and its boundary ∂D is piecewise smooth, the vector field F is of class C on R . However the orientation of the curve ∂D is clockwise (denote it by ∂D ) so the orientation + of the curve should be reversed to ∂D to use the Green’s theorem. The equation of the straight line passing through the points (2,0) and (0,−2) is y = x −2. Using now the Green’s theorem we have 2 x 2 x2 W = (∫ y + e ) dx + (3 + )y ydy = − ∫ (y + e )dx + (3x + y)ydy= ∂D ∂D+ 2 2x−x2 2 1 2 2 = − ∫∫(3y − 2y)dA = − ∫∫ ydydx = − ∫[2x − x − (x − 2) ]dx = D 0 x−2 2 0 2 2 =− 1 [2x − x − (x − 2) ]dx = − 1 (6x − 2x − 4)dx = − 1[3 x − 4 − 2 x3]2 = 4 2∫ 2 ∫ 2 3 0 3 0 0 2 2 2 3. [8 marks] Find the mass of the part of the sphere x + y + z = 2 that is cut by the paraboloid z = x + y 2(and within the paraboloid) if its density at each point is equal to the distance of the point from the z-axis. The mass density is d(x, y,z) = x + y 2 . 2 2 2 2 2 For the intersection of x + y + z = 2 and z = x + y we get the equation for z : 2 z + z − 2 0. Hence z = 1 (we neglect the other solution z = −2) and the radius of π the sphere is 2 . So we get φ = (
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