MAT2615 Assignment 4 Solutions Year Module 2026 {Sketch Available}

UNIVERSITY OF SOUTH AFRICA
Department of Mathematical Sciences


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MAT2615: Calculus in Higher Dimensions

Assignment 4 (AS4) | Year Module, 2026

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MAT2615
Module Code:
Calculus in Higher Dimensions
Module Name:
Vector Calculus: Line, Surface and Flux Inte-
Assignment Topic:
grals
AS4/0/2026
Assignment Number:
Year Module
Module Type:
43
Total Marks:




Submitted in partial fulfilment of the requirements for MAT2615, UNISA 2026

,UNISA | MAT2615 Calculus in Higher Dimensions – Assignment 4



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, so 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): 2 = 2 cos t and 0 = 2 sin t, so t = 0.

At the ending point (−2, 0): −2 = 2 cos t, so cos t = −1, hence t = π.

Therefore 0 ≤ t ≤ π.

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

so that
r′ (t) = (−2 sin t, 2 cos t), dr = (−2 sin t, 2 cos t) dt.




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,UNISA | MAT2615 Calculus in Higher Dimensions – Assignment 4


Substituting x = 2 cos t, y = 2 sin t into F(x, y) = (6xy − 12, 3x2 ):


6xy − 12 = 6(2 cos t)(2 sin t) − 12 = 24 sin t cos t − 12,

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 Z π
F(r(t)) · r′ (t) dt = 24 sin t cos t − 12, 12 cos2 t · (−2 sin t, 2 cos t) dt.


W =
0 0



Calculating the dot product:

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



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


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


First integral. Let u = sin t, so du = cos t dt:

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


since sin 0 = 0 and sin π = 0.

Second integral.

Z π
sin t dt = [− cos t]π0 = − cos π − (− cos 0) = 1 + 1 = 2.
0



Third integral. Using cos3 t = cos t(1 − sin2 t), and letting u = sin t:

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




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,UNISA | MAT2615 Calculus in Higher Dimensions – Assignment 4


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




W = 48 units


(b) Using the Fundamental Theorem of Line Integrals


We have F(x, y) = (6xy − 12, 3x2 ). Let


P (x, y) = 6xy − 12, 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 = 3x2 y − 12x + g(y),


where g(y) is a function of y.

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

Since fy = Q = 3x2 , we have 3x2 + g ′ (y) = 3x2 , so g ′ (y) = 0, hence g(y) = C. Taking g(y) = 0:


f (x, y) = 3x2 y − 12x.



The Fundamental Theorem of Line Integrals gives

Z
F · dr = f (−2, 0) − f (2, 0).
C



Calculating:
f (−2, 0) = 3(−2)2 (0) − 12(−2) = 0 + 24 = 24,

f (2, 0) = 3(2)2 (0) − 12(2) = 0 − 24 = −24.


Hence
W = 24 − (−24) = 48.



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, UNISA | MAT2615 Calculus in Higher Dimensions – Assignment 4




W = 48 units


Thus both methods give
Work done = 48 units .




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