MAT235 LEC0101 - Summer 2022
Practice Final Exam - Aug 19, 2022
Instructor: Xiao Jie
Time allotted: 180 minutes.
Aids permitted: None.
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@mail.utoronto.ca
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• DO NOT start the test until instructed to do so.
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• This test consists of 8 questions, some with multiple parts.
• The number of marks for each part of a question is indicated in the question. Total Marks: 100
• This test has 18 pages, including this cover page. Make sure you have all of them.
• JUSTIFY YOUR ANSWER FULLY IN THE SPACE PROVIDED.
Question 1. Cylindrical Coordinates and Spherical Coordinates [15 marks]
Q43 Q44 of Notes, and Q2, Q7b of Midterm 3
Question 43
p p
p
Consider the surface A characterized by z = 3 (x2 + y 2 ) = 3 · r,
where r is the radius in cylindrical coordinates.
Consider the surface B as the sphere of radius 2, x2 + y 2 + z 2 = 4.
Find the volume bounded by surface A and B.
Question 44
p
Compute the volume above the cone z = x2 + y 2 and below the sphere x2 + y 2 + z 2 = 2z.
Q2 a) Integrate f (x, y, z) = p
x2 + y 2 + z 2 over the domain D bounded by:
x2 + y 2 + z 2 = 4, below z = x2 + y 2 , above z = 0.
Q2 b) Integrate f (x, y, z) =p
x2 + y 2 + z 2 over the domain D bounded by:
2
2
2
x + y + z = 4, below z = x2 + y 2 , above z = 0.
2
Q7 b) Consider the surface created by
p
x2 + y 2 + z 2 = 1
p
z
x2
Plot the surface as a curve on the rz-plane (with only r
Find the volume enclosed by the surface.
+ y2 + z2
0).
Using Spherical coordinates, we see that the equation is given by ⇢ = 1 cos .
rotates downward on the rz-plane, which is also clockwise.
As increases from 0 to ⇡, ⇢ increases from 0 to 2, creating the heart.
We see that ⇢max is given by the blue curve ⇢ = 1 cos .
Z
Z ✓=2⇡ Z =⇡ Z ⇢=1 cos
1=
⇢2 sin d⇢ d d✓
D
✓=0
= 2⇡
=
Let u = 1
cos , du = sin d
=
⇡
(1
6
Z
2⇡
3
=0
⇢=0
=⇡
sin
=0
Z
cos )4
=⇡
⇣ ⇢3
3
⇢=1 cos
⇢=0
sin (1
⌘
d
cos )3 d
=0
Z
2⇡
=
u3
3
2⇡ u4
=
3 4
=⇡
⇡
= (24
6
=0
3
0) =
16⇡
8⇡
=
6
3
Question 2. Line Integral [10 marks]
Refer to Q57 Q58 of Notes.
Question 57
Compute the line integral.
3
3
3
2
~
a)
3y 2 /x2 )
R F (x, y) = (4x + 3y + 2y /x , 3x 3y
~ r where C is the line segment from (2, 2) to (1, 0).
C F · d~
2 2y
4x
3 2y
4x
~
b)
R F (x, y) = (3x e + 4ye , 2 + 2x e 2+ e )
~ r where C is given by ↵(t) = (3t, t ) with t 2 (1, 2).
C F · d~
2
~
c)
4y + 5, 2xy 4x 9)
R Let F (x, y) = (y
~
r where C is the upper half of the ellipse x2 /9 + y 2 /4 = 4 with the counter-clockwise
C F · d~
direction.
Question 58
Compute
the line integral.
R
2
a) C yx dx x2 dy where C is given by the left half of the circle with
radius 5 counter-clockwise, followed by the line segment from (0, -5) to (0, 5).
R
b) C (y 4 2y) dx + ( 6x + 4xy 3 ) dy where C is the boundary of the rectangle with x 2 (0, 3),
y 2 (0, 2), traversed clockwise.
3
~
c)
xy 2 , 2 x3 ).
R Let F (x, y) = (y
~ r where C is the boundary of the solid circle in the first quadrant with radius 4 (centered at
C F · d~
the origin), traversed counter-clockwise.
2
2
~
d)
R Let F (x, y) = (2y x , 7x + y ).
~ r where C is the union of 2 circles. The first circle is of radius 1, clockwise. The second
C F · d~
circle is of radius 3, counter-clockwise.
2
2 2
sin(y) ).
~
e)
R Let F (x, y) = (y x + x , x y + x e
~ r where C is the top half of the circle of radius 1 counter-clockwise.
C F · d~
Note: The curve C is not a closed loop. We can add a curve to create a loop.
4
Question 3. Surface Integral [15 marks]
Question 4. Surface Integral [15 marks]
Refer to Q62 to Q67 of Notes.
Question 62
R
~
Compute the flux of the vector field through the surface: S F~ · dS.
a) Let F~ (x, y, z) = (2x, 2y, 2z).
S = {x2 + y 2 = 9, 0  z  5} be the capless cylinder, with outward normal.
b) Let F~ (x, y, z) = (0, 0, x2 + y 2 ).
Let S be the disk of radius 3 at z = 3, with upward normal.
c) Let F~ (x, y, z) = (3x2 , 2y, 8).
Let S be the portion of the plane 2x + y + z = 0 for (x, y) 2 [0, 2] ⇥ [0, 2], with normal in the -z
direction.
d) Let F~ (x, y, z) = (z, 0, 0).
Let S be the
R triangle given by (1, 0, 0), (0, 2, 0), (0, 1, 1), with upward normal.
~
Compute S F~ · dS
Question 63
Let F~ = (3xy 2 , xez , z 3 ).
Let S be the surface of the solid bounded by the cylinder x2 + y 2 = 1 and the plane z = 1 and
z = 2, oriented
with the normal direction outward.
R
~
Compute S F~ · dS.
Question 64
Let F~ = (2ycosz, ex sinz, xey ).
Let S is the
R upper hemisphere of radius 2, with the normal direction being upward.
~
Compute S (r ⇥ F~ ) · dS.
Strategy:
Use Stokes Theorem on the boundary (curve) of the surface.
Question 65
Let F~ = (yz, 2x, 3x).
Let C be the curve of intersection by plane x + z = 5 and the cylinder x2 + y 2 = 9, oriented counterclockwise Rwhen viewed from above.
Compute C F~ · d~r using Stokes Theorem.
5
Question 66
Let F~ = (zesiny , 3x2 , xyz).
y2
Let S be the surface of the ellipsoid given by x2 +
+ z 2 = 1, with y > 0.
4
p
Let ↵(u, v) = (u, 2 1 u2 v 2 , v), be the natural parametrization of S, which determines the normal direction
R of S. What is the domain of ↵?
Compute C F~ · d~r, where C is the boundary of S, with the direction of C determined by Stokes’
Theorem.
Question 67
Let F~ = (esin(y) yz 2 , arctan(xz), 3 + y).
Let S be the surface given by z = 1 x2
outward. R
~
Compute S F~ · dS.
y 2 above the xy-plane, oriented with the normal direction
Note: the surface is not closed.
6
(extra space)
7
(extra space)
8
Question 5. [10 marks]
Refer to Q59, Q68 of Notes.
Question 59
y
x
Consider the Magnetic Field F~ (x, y) = ( 2
,
).
2 x2 + y 2
x
+
y
R
Compute C F~ · d~r. Note: F~ is not di↵erentiable at the origin.
1. Let C be a circle centered at the origin, radius R counterclockwise.
2. Let C be an arbitrary loop that does not enclose the origin.
3. Let C be an arbitrary loop, encircling the origin once counterclockwise.
Strategy: We can not use Green’s Theorem directly as F~ is not di↵erentiable at the origin. We
can add an extra small circle of radius R = ✏ around the origin, and apply the theorem on the region in betwee
4. Let C be an arbitrary curve starting from a point on the positive x-axis, rotating counterclockwise, ending at a point on the negative x-axis, with the curve always above the x-axis.
Strategy: We can try to enclose an area by adding the horizontal line on the x-axis, but we
must avoid the origin where F~ is not di↵erentiable. So we introduce a half circle of radius R = ✏
around the origin.
Question 68
1
Consider the Electric Field F~ (x, y, z) = 2
(x, y, z).
2
(x + y + z 2 )3/2
R
~
Compute S F~ · dS.
Note: F~ is not di↵erentiable at the origin.
1. Let S be a sphere centered at the origin, radius R with outward normal.
2. Let S be an arbitrary closed surface that does not enclose the origin.
3. Let S be an arbitrary surface, enclosing the origin with outward normal.
Strategy: We can not use Divergence Theorem directly as F~ is not di↵erentiable at the origin.
We can add an extra small sphere of radius R = ✏ around the origin, and apply the theorem on the region in b
4. Let S be an arbitrary dome with outward normal, that is sitting on the xy-plane, with z 0,
having a boundary C, which is a closed curve on the xy-plane.
Strategy: We can try to enclose a volume by adding the horizontal plane on the xy-plane
enclosed by C, but we must avoid the origin where F~ is not di↵erentiable. So we introduce a
half sphere of radius R = ✏ around the origin.
9
(extra space)
10
Question 6. Change of Variables (15 marks)
Integrate f (x, y, z) = z 2 over the solid ellipsoid D defined by
⇣ x ⌘2
a
+
⇣ y ⌘2
b
+
⇣ z ⌘2
c
=1
(where a, b, c are constants)
Strategy: Consider a linear transformation that would turn a unit sphere into the ellipsoid.
Then instead of integrating over the ellipsoid, we may integrate over the unit sphere.
What is g and f g for this linear transformation?
How would we integrate over the unit sphere?
Consider
the standard basis to sizes a, b, c:
2 3 the
2 linear
3 2transformation
3 2 3 that
2 stretches
3 2 3
u
u
a 0 0
u
au
x
g 4 v 5 = A 4 v 5 = 40 b 05 4 v 5 = 4 bv 5 = 4y 5.
w
w
0 0 c
w
cw
z
Det(Dg ) = Det(A) = abc
g will turn the unit sphere into the ellipsoid.
Z
Z
2
z =
c2 w2 abc
D
unit sphere
We use spherical coordinates to integrate the unit sphere in uvw-space.
We have w = ⇢ cos in spehrical coordinates (of uvw-space).
Z ⇢=1 Z =⇡ Z ✓=2⇡
3
= abc
⇢2 cos2 ⇢2 sin d✓ d d⇢
⇢=0 Z =0
✓=0
⇡
1
2
3
= abc 2⇡
cos sin d
5
0
Let u = cos
Z , du = sin d , sin d = du
2⇡
3
= abc
u2 ( 1)du
5
2⇡
u3
= abc3 ( 1)
5
3
⇡
2⇡
3
3
= abc
cos
15
0
2⇡
3
3
= abc
( 1)
13
15
4⇡
= abc3
15
11
(extra space)
12
Question 7. Change of Variables [10 marks]
Refer to Q8 of Midterm 3
Let
1 (x2 +y2 )/2
e
2⇡
a) [1 marks] Using Polar Coordinates, integrate f (x, y) over B, the solid unit circle at the origin.
f (x, y) =
b) We will try to do part a) using a di↵erent change of variables.
Consider
x
u = x2 + y 2
v=p
2
x + y2
In other words, we have defined
h(x, y) = (x2 + y 2 , p
x
x2
+ y2
) = (u, v)
p
1. [1 marks] It is clear that x = v u. Find an expression for |y| in terms of u and v.
2. [3 marks] Find Dh , and DetDh . Then, write DetDh in terms of u and v.
3. [7 marks] Using the change of variable formula presented in class, integrate f (x, y) over B by
integrating the corresponding region A in the uv-plane.
Strategy: To find A, first understand
Z what u and v represent in the xy-plane.
1
p
Also, it may be useful to know that
= arcsin x
1 x2
4. [8 marks] Did you get the same answer as part a)? Explain.
a)
Z
f (x, y) =
B
Z
r=1 Z ✓=2⇡
r=0
=
=
(e
✓=0
e
1/2
r 2 /2
1
e
2⇡
r 2 /2
r d✓dr
r=1
r=0
1) = 1
e
1/2
b)
u = x2 + y 2 = v 2 u + y 2
y2 = u
"
v 2 u = u(1 v 2 )
p
|y| = u(1 v 2 )
2x
2y
Dh =
1
1
(x2 + y 2 ) 1/2 + x( )(x2 + y 2 ) 3/2 (2x) x( )(x2 + y 2 )
2
2

2x
2y
=
2
2
1/2
2
2
2
3/2
2
(x + y )
x (x + y )
xy(x + y 2 ) 3/2
13
3/2 (2y)
#
2x2 y(x2 + y 2 )
Det(Dh ) =
⇣
2y (x2 + y 2 )
3/2
=
x2
We see that,
u = r2 , so for the unit circle, u : 0 ! 1.
x
v = = cos ✓, so for the unit circle, v = cos ✓ :
r
p
1
x2 + y 2
Note that Det(Dg ) =
=
Det(Dh )
2y
Z
f (x, y) =
B
Z
u=1 Z v=1
u=0
x2 (x2 + y 2 )
2y
+ y2
3/2
⌘
1!1
v= 1
p
p
x2 + y 2
u
1
Note that
=
= p
2y
2y
2 1 v2
Z u=1 Z v=1
1
=
e
u=0
v= 1 2⇡
1/2
1
e
2⇡
u/2
p
x2 + y 2
dv du
2y
1
dv du
2 1 v2
u=1 ⌘⇣
v=1 ⌘
1 ⇣ u/2
=
e
( 2)
arcsin v
4⇡
u=0
v= 1
⇣
⌘⇣
⌘
1
⇡
⇡
=
e 1/2 1
2⇡
2
2
1
=
(1 e 1/2 )⇡
2⇡
1
= (1 e 1/2 )
2
1
We see that the calculations in part b) yielded an extra factor of .
2
This is because h(x, y) was not a one to one function, since 2 points of (x, y) would have the same
cos ✓ value.
Indeed, for (x, y) = (0, 1), and (x, y) = (0, 1), we have h(x, y) = (1, 0).
This is not good since for g to be the inverse of h, we would need to set g(1, 0) = (0, 1) and
g(1, 0) = (0, 1), which is not possible, as g must be a function so g can only have a unique output
for the same input.
u/2
p
In order to rectify this issue, to integrate over B properly, one would need to split B into the
upper half circle, and the lower half circle.
First, define g1 to transform the square A on the uv-plane, to the upper half of the circle on
the xy-plane, by setting y to be the positive square root value. This is what is done via the calcula1
tion above, which yields (1 e 1/2 ).
2
Then, define g2 to transform the square A on the uv-plane, to the bottom half of the circle on
the xy-plane, by setting y to be the negative square root value. This is also what is done via the
14
calculation above, since regardless of y being the positive or the negative square root, the calculations
1
were the same, since we had |y| in the integral, which yields (1 e 1/2 ).
2
Thus, the integral over B, would be the sum of integrals over the upper half and the bottom half,
1
which is done via the transformations g1 and g2 , each yielding (1 e 1/2 ), so the sum is exactly
2
(1 e 1/2 ) as in part a).
15
Question 8. (10 marks)
...
16
(extra space)
17
(extra space)
18