MAT2615 Assignment 4 2026 Solutions Year Module 2026 {Sketch Drawn}

UNIVERSITY OF SOUTH AFRICA (UNISA)
Department of Mathematical Sciences



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CALCULUS IN HIGHER DIMENSIONS
Assignment 4 (AS4)

Year Module — 2026


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Module Code: MAT2615

Module Name: Calculus in Higher Dimensions

Assignment No.: Assignment 4 (AS4/0/2026)

Semester: Year Module, 2026

Total Marks: 43




Submitted in partial fulfilment of the requirements for MAT2615
at the University of South Africa.

,UNISA | MAT2615 Assignment 4 — Calculus in Higher Dimensions



Question 1

Consider the force field
F(x, y) = (6xy − 12, 3x2 ).


Use the formula for a line integral to determine the work done by the force field F in moving
an object in an anticlockwise direction from the point (2, 0) to the point (−2, 0) along the
circle
x2 + y 2 = 4

by applying:

(a) The method for evaluating a line integral described in Example 16.4.1.

(b) The Fundamental Theorem of Line Integrals.


(a) Using the method for evaluating a line integral


The work done is
Z
W = F · dr.
C


The curve is
x2 + y 2 = 4.


The radius is
r = 2.


A suitable parametrization of the circle is


x = 2 cos t, y = 2 sin t.



Since the motion is anticlockwise, t increases.

At the starting point (2, 0),
x = 2, y = 0.


Therefore,
2 = 2 cos t

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,UNISA | MAT2615 Assignment 4 — Calculus in Higher Dimensions


and
0 = 2 sin t.


Thus,
t = 0.


At the ending point (−2, 0),
x = −2, y = 0.


Therefore,
−2 = 2 cos t.


Hence,
cos t = −1,

so
t = π.


Therefore,
0 ≤ t ≤ π.


The position vector is
r(t) = (2 cos t, 2 sin t).


Differentiate:
r′ (t) = (−2 sin t, 2 cos t).


Therefore,
dr = (−2 sin t, 2 cos t) dt.


The force field is
F(x, y) = (6xy − 12, 3x2 ).


Substitute
x = 2 cos t, y = 2 sin t.




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,UNISA | MAT2615 Assignment 4 — Calculus in Higher Dimensions


Then
6xy − 12 = 6(2 cos t)(2 sin t) − 12

= 24 sin t cos t − 12.


Also,
3x2 = 3(2 cos t)2

= 12 cos2 t.


Thus,
F(r(t)) = (24 sin t cos t − 12, 12 cos2 t).


Now,
Z π
W = F(r(t)) · r′ (t) dt.
0


Therefore,
Z π
W = (24 sin t cos t − 12, 12 cos2 t) · (−2 sin t, 2 cos t) dt.
0


Calculate the dot product:

Z π
(24 sin t cos t − 12)(−2 sin t) + (12 cos2 t)(2 cos t) dt.
 
W =
0



Expand:
Z π
−48 sin2 t cos t + 24 sin t + 24 cos3 t dt.
 
W =
0


Therefore,
Z π Z π Z π
2
W = −48 sin t cos t dt + 24 sin t dt + 24 cos3 t dt.
0 0 0


For the first integral, let
u = sin t.


Then
du = cos t dt.




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,UNISA | MAT2615 Assignment 4 — Calculus in Higher Dimensions


Thus, π
π
sin3 t
Z 
2
sin t cos t dt = .
0 3 0


Since
sin 0 = 0, sin π = 0,

we get
Z π
sin2 t cos t dt = 0.
0


For the second integral,
Z π
sin t dt = [− cos t]π0 .
0


Therefore,
= − cos π − (− cos 0)

=1+1

= 2.


For the third integral,
Z π
cos3 t dt.
0


Using
cos3 t = cos t(1 − sin2 t),

we have
Z π Z π
cos3 t dt = cos t(1 − sin2 t) dt.
0 0


Let
u = sin t, du = cos t dt.


Therefore, π
π
sin3 t
Z 
3
cos t dt = sin t − .
0 3 0


Hence,
=0−0

= 0.


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, UNISA | MAT2615 Assignment 4 — Calculus in Higher Dimensions


Therefore,
W = −48(0) + 24(2) + 24(0).


Thus,
W = 48 .


Therefore, the work done is
48 units .


(b) Using the Fundamental Theorem of Line Integrals


We have
F(x, y) = (6xy − 12, 3x2 ).


Let
P (x, y) = 6xy − 12

and
Q(x, y) = 3x2 .


To find a potential function f (x, y), integrate P with respect to x:

Z
f (x, y) = (6xy − 12) dx.



Therefore,
f (x, y) = 3x2 y − 12x + g(y),

where g(y) is a function of y.

Differentiate with respect to y:
fy (x, y) = 3x2 + g ′ (y).


Since
fy = Q = 3x2 ,

we have
3x2 + g ′ (y) = 3x2 .



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