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Past exam of the mathematics course of the University of Cambridge
/
2024
/
ia
/
Paper 3
/
11B
/
iv
/
Solution
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Past exam of the mathematics course of the University of Cambridge
2024
ia
Paper 3
11B
iv
Created
2026-09-24
Updated
2026-09-25
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The
first
field
is
B
1
=
∇
(
e
x
2
cos
(
yz
)
)
,
(1)
so its
integral
around
C
is zero and it is conservative with potential
e
x
2
cos
(
yz
)
.
Along
C
,
d
r
=
(
−
sin
s
,
cos
s
,
0
)
d
s
and
B
2
=
(
−
sin
s
,
cos
s
,
0
)
,
(2)
so
∮
C
B
2
⋅
d
r
=
∫
0
2
π
1
d
s
=
2
π
.
(3)
Thus
B
2
is not conservative and cannot be
a
gradient
on the stated punctured region.
Finally,
B
3
=
∇
lo
g
(
x
2
+
y
2
+
z
2
)
.
(4)
Its
integral
around
C
is zero, and it is conservative with the displayed potential. Therefore the three
integrals
are respectively
0
,
2
π
,
0
,
(5)
and exactly
B
1
,
B
3
are conservative
gradient
fields
.
Ancestors
(11)
iv
11B
Paper 3
ia
2024
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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