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This is the Exam of Multivariate Calculus and its key important points are: Parallel, Plane Containing the Points, Equations, Parametric Equations, Curve, Unit Vector, Directional Derivative, Maximum Value, Circle, Critical Points
Typology: Exams
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Name Instructor
Signature Recitation Instructor
Div. Sect. No.
spaces.
section 03, fill in 0203 and blacken the corresponding circles, including the circles for
the zeros. (If you do not know your division and section number ask your instructor.)
swer in the spaces provided for questions 1–20. Do all your work on the question sheets.
Turn in both the mark-sense sheets and the question sheets when you are finished.
may be considered if your grade is on the borderline.
back of the test pages for scratch paper.
the line whose equations are x = y, z = 0.
A. x + y = 1
B. x − y = 1
C. x − y + z = 1
D. x−y +2z = 1
E. x + y − z = 1
t cos t~i + e
t sin t~j be parametric equations of a curve C. Find the length of
C from t = 0 to t = π.
2 e
π
2 (e
π −e
−π )
2 (e
π − 1)
2 (e
π
+e
−π
)
2 (e
π
2
2 = 5 is:
3
3
critical points are
A. both maximum points
B. both minimum points
C. one minimum and one maximum point
D. one minimum and one saddle point
E. one maximum and one saddle point
√
π
0
π
y
2
(sin x
2
)dx dy?
π
0
π
√
x
(sin x
2
)dy dx
π
0
√
x
0
(sin x
2
)dy dx
√
π
√
x
π
0
(sin x
2
)dy dx
π
0
√
π
x
(sin x
2
)dy dx
√
π
0
π
x
(sin x
2
)dy dx
2
2
= 1 and above the line y = x,
then
R
x dA expressed in polar coordinates is:
∫ 3 π
2
π
2
1
0
r cos θdr dθ
∫ 3 π
2
π
2
1
0
r
2
cos θdr dθ
∫ 5 π
4
π
4
r cos θ
0
r
2
cos θdr dθ
∫ 5 π
4
π
4
r
2
0
r
2
cos θdr dθ
∫ 5 π
4
π
4
1
0
r
2
cos θdr dθ
1 − x
2
− y
2
and below by the xy-plane, the volume of D is:
π
π
π
π
π
√
2
0
√
2 −x
2
−
√
2 −x
2
8 −x
2 −y
2
3 x
2 +3y
2
xyz dz dy dx in cylindrical coordinates is:
∫ π
2
0
√
2
0
√
8 −r
2
√
3 r
r
3
z cos θ sin θdz dr dθ
∫ π
2
0
2
0
√
8 −r
2
√
3 r
r
2
z cos θ sin θdz dr dθ
∫ π
2
−
π
2
√
2
0
√
8 −r
2
√
3 r
r
3
z cos θ sin θdz dr dθ
∫ π
2
−
π
2
2
0
√
8 −r
2
√
3 r
r
3
z cos θ sin θdz dr dθ
∫ π
2
−
π
2
2
0
√
8 −r
2
√
3 r
r
2
z cos θ sin θdz dr dθ
below by the upper nappe of the cone z
2
= x
2
2
with mass density at each point
equal to x
2
2
2 is given by:
2 π
0
∫ π
4
0
2
0
ρ
2
sin φ dρ dφ dθ
2 π
0
∫ π
4
0
2 sec φ
0
ρ
4
sin φ dρ dφ dθ
2 π
0
∫ π
4
0
2
0
ρ
4
sin φ dρ dφ dθ
2 π
0
∫ π
4
−
π
4
2 sec φ
0
ρ
3
sin
2
φ dρ dφ dθ
2 π
0
∫ π
4
0
2 sec φ
0
ρ
3
sin
2
φ dρ dφdθ
F (x, y, z) = (z + y
2 )
i + 2xy~j + (x + y)
k. Find the curl(
F ) at the point (1, 1 , 1).
i +
j +
k
j
i
i
j
F (x, y) = (2xe
y
i + (x
2 e
y )
j is a conservative vector field, that is
F (x, y) = grad f (x, y) for some function f , and C is any smooth curve from (0, 0) to
(1, 1) then
C
F · d~r =
C. 2 e
D. e + 1
E. 2 e + 1
2 and the y-axis. If C is the boundery
of the region R oriented counterclockwise, then
C
(e
2 x
2
)dx + (14xy + y
2
)dy =
Σ
4 z dS, where Σ is the part of the sphere x
2
+y
2
+z
2
=
10 which lies above the plane z = 1.
10 π
10 π
10 π
D. 19 π
E. 99 π
x
2
2 and the plane z = 1 whose
boundary surface Σ is oriented by the unit normal ~n directed outward from D. If
F (x, y, z) = (8xz)
i + (z
3 e
−x )
j + (y cos x)
k then
Σ
F · ~n dS =
B. π
π
D. 2 π
16 π