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Engr 233 final 2022 exam
Applied Advanced Calculus (Concordia University)
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Concordia University
Gina Cody School of Engineering and Computer Science
Applied Advanced Calculus - ENGR233/4 (Sections J, R, S, T, U, V, X)
2022 Regular Final Examination
April 21, 2022 (12:00 PM – 3:00 PM)
1- 10%] Find parametric equations for the tangent line to the curve of intersection of the
cylinder x2 + y2 = 2 and the paraboloid z = 2x2 + 3y2 at the point (1, 1, 5).
Solution:
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2- The function z(x, y) is implicitly defined by the equation x3 z2 + 3 xyz3 + x2 y4 = 4.
(a) [7.5%] Find the partial derivatives ∂z/∂x and ∂z/∂y in terms of x, y, and z.
(b)[7.5%] Write equations of the tangent plane and the normal line to the surface
x3 z2 + 3 xyz3 + x2 y4 = 4 at the point (1, 1, 1).
Solution:
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3- [10%] Verify that the vector field F =2 xy 2i + ( 2 x 2 y + 4 y ) j is conservative and evaluate the
P2
integral
 F  dr , where coordinates of points P
1
and P2 are (0, 0) and (3, 4) , respectively
P1
Solution:
The integral is path independent if the vector field is conservative. In that case we can find a
potential function  for F . F = 2 xy 2i + ( 2 x 2 y + 4 y ) j = Pi + Qj
P
= 4 xy
y
and
Q
= 4 xy
x
The vector field is conservative if and only if
such that    = F . That is,

and
=P
x
P Q
. Then there exists a Potential function 
=
y x

=Q
y
A Potential function  that generates the vector field F = 2 xy 2i + ( 2 x 2 y + 4 y ) j can then be
obtained as
 = x2 y 2 + 2 y 2 + c
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(3,4)

F  dr =  (3, 4) −  (0, 0) = 176
( 0,0 )
4- [10%] Evaluate the double integral over the given region; ∬R xy2 dA , where R is
bounded by x + y + 1=0 and x + y2 = 1
Solution:
Chart Title
4
2
0
-5
-4
-3
-2
-1
0
1
2
3
-2
-4
2
∫ ∫
1−𝑦2
−1 −1−𝑦
1 2 1 − 𝑦2 2
1 2
𝑥 |
𝑦 𝑑𝑦 = ∫ {(1 − 𝑦 2 )2 − (−1 − 𝑦)2 }𝑦 2 𝑑𝑦
−1 − 𝑦
2 −1
−1 2
𝑥𝑦 2 𝑑𝑥𝑑𝑦 = ∫
2
1 2 6
1 𝑦 7 3𝑦 5 𝑦 4 2
∫ 𝑦 −3𝑦 4 − 2𝑦 3 𝑑𝑦 = (
−
− | )
2 −1
5
2 −1
2 7
1 1280 − 1344 − 560 + 10 − 42 + 35
−621
= (
)=
= −4.44
2
70
140
5- [15%] Use Green theorem to evaluate the line integral
2
2
∮ (2 x y ex y + 3x2 y) dx+ (x2 e x y ) dy,
C
where C is the ellipse x2 + 4 y2 = 4
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Solution:
∮𝐶(2𝑥𝑦𝑒 𝑥
2𝑦
P = 2𝑥𝑦𝑒 𝑥
Q=𝑥 2 𝑒 𝑥
2𝑦
2
+ 3𝑥 2 𝑦) 𝑑𝑥 + (𝑥 2 𝑒 𝑥 𝑦 )𝑑𝑦 =∬𝑅
2𝑦
+ 3𝑥 2 𝑦
𝜕𝑃
𝜕𝑄
𝜕𝑦
𝜕𝑄 𝜕𝑃
−
= −3𝑥 2
𝜕𝑥
𝜕𝑦
𝜕𝑦
= 2𝑥𝑒 𝑥
= 2𝑥𝑒 𝑥
2𝑦
2𝑦
y= v
We calculate the Jacobean =
𝜕𝑥
|𝜕𝑢
𝜕𝑦
𝜕𝑢
𝜕𝑥
𝜕𝑥
𝜕𝑣
|
𝜕𝑦
𝜕𝑣
−
𝜕𝑃
𝜕𝑦
+ 2𝑥 3 𝑦𝑒 𝑥
+ 2𝑥 3 𝑦𝑒 𝑥
The region of the integration R is an ellipse
Let x=2u
𝜕𝑄
𝑥2
4
𝑑𝐴
2𝑦
2𝑦
+ 3𝑥 2
+ 𝑦2 = 1
2 0
|=2
=|
0 1
The new region R1 is a circle with a radius of 1
u= r cos𝜃 v= r sin𝜃 dudv= rdrd𝜃
∬
𝑅
𝜕𝑄 𝜕𝑃
−
𝑑𝐴
𝜕𝑥
𝜕𝑦
= ∬ −3𝑥 2 𝑑𝐴 = ∬ −3(2𝑢)2 (2)𝑑𝑢𝑑𝑣 = ∫
𝑅
𝑅1
2𝜋
0
1
∫ −24 𝑟 2 𝑐𝑜𝑠 2 𝜃 𝑟𝑑𝑟 𝑑𝜃
0
2𝜋
2𝜋
1
1
2𝜋
= -6∫0 𝑟 4 | 𝑐𝑜𝑠 2 𝜃 𝑑𝜃 = −3 ∫0 (1 + 𝑐𝑜𝑠2𝜃)𝑑𝜃 = −3 (𝜃 + 𝑠𝑖𝑛2𝜃 | ) = −3( 2𝜋 + 0) =
2
0
0
−6𝜋
6- [15%] Compute the line integral
∮ (2xy − x2 + cosx) dx + (x + y2 − sin y ) dy
C
where C is the closed curve bounding the region of the plane between y = x2 and x = y2,
with clockwise orientation.
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7- [10%] Compute the following iterated triple integral
0
∫ (∫
−1
√1−x2
− √1−x2
(∫
√1−x2 −y2
− √1−x2 −y2
1
( x 2 + y 2 + z 2 ) ⁄3 dz) dy) dx
Solution:
8- [15%] Use the divergence theorem to find the outward flux of the vector field:
𝐹⃗ = xy2 ⃗⃗i + x2y ⃗j + 6 sin x k⃗
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through the boundary of the solid bounded by the cone 𝑧 = √x2 + y2 and the planes z = 1
and z = 2.
Hint: use spherical coordinates.
Solution:
Adjust the breakdown on the basis of a total of 15.
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