18.02 Practice Exam 3 A – Solutions
1. a) The area of the triangle is 2, so ȳ =
1
1
2
2−2y
y dxdy.
0
2y−2
b) By symmetry x̄ = 0.
r2 δ rdrdθ =
2. δ = |x| = |r cos θ|. I0 =
D
2π
1
π/2
1
r
2 |r cos θ|rdrdθ = 4
0
0
π/2
r4 cos θdrdθ = 4
0
0
0
1
4
cos θdθ =
5
5
3. a) Nx = 6x2 + by 2 , My = ax2 + 3y 2 . Nx = My provided a = 6 and b = 3.
b) fx = 6x2 y + y 3 + 1 =⇒ f = 2x3 y + xy 3 + x + c(y). Therefore, fy = 2x3 + 3xy 2 + c/ (y).
Setting this equal to N , we have 2x3 + 3xy 2 + c/ (y) = 2x3 + 3xy 2 + 2 so c/ (y) = 2 and c = 2y. So
f = 2x3 y + xy 3 + x + 2y (+constant).
c) C starts at (1, 0) and ends at (−eπ , 0), so
FF · dFr = f (−eπ , 0) − f (1, 0) = −e−π − 1.
C
1
C
1
x2 x3 dx + (x2 )2 (2xdx) =
yx3 dx + y 2 dy =
4.
0
3x5 dx = 1/2.
0
2x/y −x2 /y 2
u uy
2
=
5. a)
x
= 3x
/y. Therefore,
vx vy y
x
dudv = (3x2 /y)dxdy = 3u dxdy
4
=⇒
dxdy =
1
dudv.
3u
5
4
1
1
2
dudv =
ln 5 dv = ln 5.
3
2
1 3u
2 3
6. a)
M dx =
−My dA.
b)
C
R
1
b) We want M such that −My = (x + y)2 . Use M = − (x + y)3 .
3
3
1
x
1
1
7. a) div FF = 2y, so
2y dA =
2y dydx =
x6 dx = .
7
R
0
0
0
b) For the flux through C1 , n̂ = −ĵ implies FF · n̂ = −(1 + y 2 ) = −1 where y = 0. The length of
C1 is 1, so the total flux through C1 is −1.
The flux through C2 is zero because n̂ = ı̂ and FF ⊥ ı̂.
1
8
div FF dA −
FF · n̂ds =
FF · n̂ds −
FF · n̂ds = − (−1) − 0 = .
c)
7
7
C3
R
C1
C2
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18.02SC Multivariable Calculus
Fall 2010
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