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Past exam of the mathematical methods II course of the University of Oxford
/
2018
/
9
/
c
/
Solution
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(
@codex,
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)
...
Year 1 of the physics course of the University of Oxford
Mathematical methods II course of the University of Oxford
Past exam of the mathematical methods II course of the University of Oxford
2018
9
c
2026-10-25
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Inside the unit
ball
,
∇
⋅
J
=
0
. On its
sphere
,
J
⋅
r
=
x
y
−
x
y
=
0
, so the
jump
to zero outside adds no
surface
divergence
. The
field
therefore satisfies the required
weak derivative
condition.
By reflection
symmetry
,
integrals
of
x
y
,
x
z
,
yz
over the
ball
vanish. Rotational
symmetry
and
spherical coordinates
give
∫
B
1
x
2
d
V
=
3
1
∫
B
1
r
2
d
V
=
3
4
π
∫
0
1
r
4
d
r
=
15
4
π
.
(1)
Hence the left side is
∫
x
(
y
,
−
x
,
0
)
d
V
=
(
0
,
−
4
π
/15
,
0
)
. Meanwhile
J
×
r
=
(
−
x
z
,
−
yz
,
x
2
+
y
2
)
, so its
integral
is
(
0
,
0
,
8
π
/15
)
. Therefore
2
1
(
1
,
0
,
0
)
×
(
0
,
0
,
8
π
/15
)
=
(
0
,
−
4
π
/15
,
0
)
,
(2)
exactly the same
vector
.
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(12)
c
9
2018
Past exam of the mathematical methods II course of the University of Oxford
Mathematical methods II course of the University of Oxford
Year 1 of the physics course of the University of Oxford
Physics course of the University of Oxford
Undergraduate course of the University of Oxford
Course of the University of Oxford
University of Oxford
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