Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
NO CALCULATORS, BOOKS, OR PAPERS ARE ALLOWED. Use the back of the test pages for scrap paper. Points awarded. 1. (5 pts). 9. ( ...
Typology: Exercises
1 / 11
This page cannot be seen from the preview
Don't miss anything!
MA 174: Multivariable Calculus
Final EXAM (practice)
NAME Class Meeting Time:
NO CALCULATORS, BOOKS, OR PAPERS ARE ALLOWED. Use the back
of the test pages for scrap paper.
Points awarded
Total Points:
Surface Integral:
If R is the shadow region of a surface S defined by the equation f (x, y, z) = c, and
g is a continuous function defined at the points of S, then the integral of g over
S is the integral
S
g(x, y, z) dσ =
R
g(x, y, z)
|∇f |
|∇f · p|
dA,
where p is a unit vector normal to R and |∇f · p| 6 = 0.
Green’s Theorem:
C
P dx + Q dy =
R
∂x
∂y
dA
where C is a positively oriented simple closed curve enclosing region R, and P ,
Q have continuous partial derivatives.
Divergence Theorem:
S
F · dS =
S
F · n dσ =
D
∇ · F dV
where D is a simple solid region with boundary S given outward orientation,
and component functions of F have continuous partial derivatives.
Stokes’ Theorem: ∮
C
F · dr =
S
∇ × F · n dσ
where C, given counterclockwise direction, is the boundary of oriented surface
S, n is the surface’s unit normal vector and component functions of F have
continuous partial derivatives.
2
2
2
= 2 are
2
+y
2
subject to the constraint x
2
− 2 x+y
2
− 4 y = 0.
normal to the tangent plane of
4 x + y
2
3 = 8
at P.
A. x = t + 4, y = t, z = −t
B. x = 4t + 2, y = 2t + 1, z = 3t − 1
x − 2
y − 1
z − 1
x − 4
y − 3
E. x = 4t − 2 , y = 2t − 1 , z = − 3 t + 1
2
xy
2
3
= 2 at the point (− 1 , 1 , 1) is:
i − 2
j + 3
k
j +
k
k
j +
k
i + 2
j + 3
k
t and y = t
2
∂z
∂x
= 2xy
2
− y
and
∂z
∂y
= 2x
2
y − x, find
dz
dt
when t = 0.
2
2
= x.
A. ρ = sin φ cos θ
B. ρ sin φ = sin
2 φ cos θ
C. ρ = sin φ cos φ
D. ρ
2 = ρ cos φ
E. ρ
2 sin
2
φ = ρ sin φ cos θ
2
. If (0, b, 5) is a point on the tangent plane to
S at (0, 1 , 2) on S, then b =
2 and x + y = 0
F = ∇f, f =
x
2
2
. If C is any smooth curve joining the points
(1, 1), (2, 2), then
C
F · d~r =
2
2 = 4, y = 1, y = 0,
and S be the boundary of D. If
F (x, y, z) =
1
3
(x
3 ~ i + y
3 ~ j + z
k), then with ~n
being the unit outward normal, evaluate
S
F · ~ndσ.
A. 8 π
π
C. 28 π
D. 10 π
angular coordinates to spherical coordinate
3
0
√
9 −x
2
0
x
2 +y
2
0
ydzdydx =
π/ 2
0
π/ 2
a
3 csc ϕ
0
bdρdϕdθ.
A. a = 0, b = ρ
2 sin ϕ
B. a = π/ 4 , b = ρ
3
sin ϕ sin θ
C. a = π/ 4 , b = ρ
3 sin
2
ϕ sin θ
D. a =
π
3
, b = ρ
3 sin
2
ϕ sin θ
E. a = −π/ 2 , b = ρ
3
sin
2
ϕ
F = 2xy~i + (x
2
2 )
j is a conservative vector field, i.e.,
F = ∇f. If f (0, 0) = 0,
then f (1, 1) =
S
ydS, where S is the part of the plane x + 2y + z = 1 in the 1st
octant.
F (x, y, z) = xz~i + xyz ~j − y
k, then curl
F evaluated at (1, 1 , 1) equals
i −
j +
k
i +
j −
k
i +
j −
k
i +
j +
k
i −
j + 2
k
2
0
2
x
e
y
2
dydx.
A. 2(e
4 − 1)
B. e
4 − 1
e
4
e
4 − 1
E. e
4
F (x, y, z) = (x sin x + y)
i + xy~j + (yz + x)
k, then curl
F evaluated at (π, 0 , 2)
equals
A. π~i −
j +
k
i −
j −
k
i − π~j +
k
i −
j + π
k
i +
j +
k
C
(2x + yz)dx + (2y + xz)dy + xydz
where c : ~r(t) = t
2
(1 + t)
i + cos
π
t
2
j +
t
2
t
4
k, 0 ≤ t ≤ 1.
S
(x
2
2
2
)dS where S is the upper hemisphere of x
2
2
2 = 2.
A. 12 π
B. 8 π
C. 6 π
D. 4 π
E. 3 π
C
2 y
x
2
2
dx +
2 x
x
2
2
dy where C is the circle x
2
2 = 1 oriented
counterclockwise.
A. 2 π
B. 4 π
D. − 4 π
E. − 2 π
S
F · ~n dS where S is the sphere x
2
2
2 = 2
oriented by the outward normal and
F (x, y, z) = 5x
i + 5y
j + 5z
k.
2 π
B. 16 π
C. 24 π
2 π
E. 20 π
cylindircal coordinates (r, θ, z) = for the point (x, y, z) =
Answer: (ρ, ϕ, θ) = (
3 , cos
− 1 (
1 √
3
π
4
Answer: (r, θ, z) = (
π
4
