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# Additional Exercises 4

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OneClass2138

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University of Toronto St. George

Mathematics

MAT237Y1

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Fall

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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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