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Question 4:
Let P ? x 2 ? 2 xy and Q ? x 2 y ? 3 . By Greens theorem, we have
? ?Q
?P ?
? Pdx ? Qdy ? ?? ?? ?x ? ?y ?? dA
D
? ?
?
?
? ?? ? ? x 2 y ? 3 ? ? ? x 2 ? 2 xy ? ? dA
?x
?y
?
D ?
? ?? ? y ? 2 x ? ? ? ?2 x ? ? dA
D
? ?? 2 x ? y ? 1? dA
D
where D is the region enclosed by y 2 ? x and x ? p . The region D can be described as
y 2 ? x ? p and ? p ? y ?
p . Hence, the double integral is calculated by:
?? 2 x ? y ? 1? dA ?
D
p
p
? ? 2 x ? y ? 1? dxdy
? p y2
p
?
? ? y ? 1? x
2 x? p
? p
x? y2
p
?
? ? y ? 1? ? p
2
dy
? y 4 ? dy
? p
p2 y2
y6 y5
?
? p2 y ? ?
2
6
5
?
p
? p
8 p 5/2
5
Now, we compute the line integral. Let C1 be the portion of the curve y 2 ? x and let C2
be the portion of the line x ? p . Therefore,
? Pdx ? Qdy ? ? Pdx ? Qdy ? ? Pdx ? Qdy
C
C1
C2
For C1 : y 2 ? x , we parameterize x ? t 2 , y ? t where ? p ? t ?
p . Therefore,
? Pdx ? Qdy ?
C1
p
? ?t
4
? 2t 3 ? 2tdt ? ? t 5 ? 3? dt
? p
??
? 2t ? t 4 ? 2t 3 ? ? ? t 5 ? 3? ? dt
?
?
p
? p
? t 6 4t 5
?
?? ?
? 3t ?
?2 5
??
?6 p?
p
p
5/2
8p
5
For C2 : x ? p , we have
? Pdx ? Qdy ? ? Qdy ? ?
p
? p
C2
? p 2 y ? 3? dy ?
C2
16 p5/2
?6 p
5
Finally,
? Pdx ? Qdy ? ? Pdx ? Qdy ? ? Pdx ? Qdy
C
C1
C2
?
? 8 p 5/2
8 p ? ? 16 p 5/2
? ?6 p ?
?
?
6
p
? ?
??
5 ? ? 5
5
?
?
5/2
Hence, the Greens theorem in plane is verified as
? ?Q ?P ?
8 p 5/2
? Pdx ? Qdy ? ??D ?? ?x ? ?y ?? dA ? 5
Question 6:
For given F , we have
?P ?Q ?R
?
?
?x ?y ?z
?
?
?
? ? 3 z 2 y 3 ? ? ? pxy 2 z ? ? ? pz ? pxyz 2 ?
?x
?y
?z
divF ?
? 0 ? px ? 2 y ? z ? ? p ? pxy ? 2 z ? ?
?p
(i) The solid bounded by paraboloid z ? 2 ? x 2 ? y 2 and plane z ? 1 can be sketched as
follow:
The triple volume integral for divF is calculated by
??? divFdV ? ??? pdV
V
V
2
2 ? z 2?
1
0
??
? ? prd? drdz
2
0
pr 2
? 2? ?
2
1
2
? 2? ?
1
r ? 2? z
dz
r ?0
p ?2 ? z?
dz
2
2
? z2
?
p?
? ? ? ? p ? ? ? 2z ? ?
2
? 2
?1
Let S1 be the surface of z ? 2 ? x 2 ? y 2 and let S 2 be the surface of plane z ? 1 that is
boundary of the solid. We have
?? F ? ndS ? ?? F ? ndS ? ?? F ? ndS
S
S1
S2
For S1 : z ? 2 ? x 2 ? y 2 , let g ? x, y ? ? 2 ? x 2 ? y 2 we have
?? F ? ndS ? ?? ? ? Pg
S1
x
? Qg y ? R ? dA
D
? ?? ?? ? ? 3 z 2 y 3 ? ? ?2 x ? ? pxy 2 z ? ?2 y ? ? ? pz ? pxyz 2 ? ?? dA
D
?
?
?
?
2
2
? ?? ? ? 3 ? 2 ? x 2 ? y 2 ? y 3 ? ?2 x ? ? pxy 2 ? 2 ? x 2 ? y 2 ? ? ?2 y ? ? p ? 2 ? x 2 ? y 2 ? ? pxy ? 2 ? x 2 ? y 2 ? ? dA
?
??
?
D
??
p?
2
where D is the planar disk x 2 ? y 2 ? 1on xy-plane.
For S 2 : z ? 1 , let z ? g ? x, y ? , we have n ? 1,0,0 and
?? F ? ndS ? ?? ? ? Pg
S1
x
? Qg y ? R ? dA
D
? ?? ? ? P ? 0 ? ? Q ? 0 ? ? R ? dA
D
? ?? ? pz ? pxyz 2 ? dA ? ?? p ?1 ? xy ? dA
D
D
2? 1
?
? ? p ?1 ? r cos? ? r sin ? ? rdrd? ? p?
0 0
Therefore,
?
?? F ? ndS ? ?? F ? ndS ? ?? F ? ndS ? ?? ?
S
S1
S2
p?
2
p?
?
? ? p? ?
2
?
Hence, we have verified the Gauss divergence theorem as
?? F ? ndS ? ??? divFdV ?
S
V
p?
2
(ii) The solid bounded by sphere 1 ? z 2 ? x 2 ? y 2 and plane z ? 0 can be sketched as
follow:
The triple volume integral for divF is calculated by
??? divFdV ? ??? pdV
V
V
1
?
2
1
?
2
p ? Volume of the sphere
2 p?
3?
?4
p ? ? ?1? ? ?
3
?3
?
Let S1 be the surface of 1 ? z 2 ? x 2 ? y 2 and let S 2 be the surface of plane z ? 0 that is
boundary of the solid. We have
?? F ? ndS ? ?? F ? ndS ? ?? F ? ndS
S
S1
S2
For S1 : 1 ? z 2 ? x 2 ? y 2 , let z ? g ? x, y ? ? 1 ? x 2 ? y 2 we have
?? F ? ndS ? ?? ? ? Pg
S1
x
? Qg y ? R ? dA
D
? ?? ?? ? ? 3 z 2 y 3 ? ? ? x ? ? pxy 2 z ? ? y ? ? ? pz ? pxyz 2 ? ?? dA
D
?
?
?
?
2
? ?? ? ? 3 ?1 ? x 2 ? y 2 ? y 3 ? ? x ? ? pxy 2 ?1 ? x 2 ? y 2 ? ? ? y ? ? p 1 ? x 2 ? y 2 ? pxy ?1 ? x 2 ? y 2 ? ? dA
??
??
D
??
p?
3
where D is the planar disk x 2 ? y 2 ? 1on xy-plane.
For S 2 : z ? 0 , let z ? g ? x, y ? , we have n ? 1,0,0 and
?? F ? ndS ? ?? ? ? Pg
S1
x
? Qg y ? R ? dA
D
? ?? ? ? P ? 0 ? ? Q ? 0 ? ? R ? dA
D
? ?? ? pz ? pxyz 2 ? dA ? ?? p ?1 ? xy ? dA
D
D
2? 1
?
? ? p ?1 ? r cos? ? r sin ? ? rdrd? ? p?
0 0
Therefore,
?
?? F ? ndS ? ?? F ? ndS ? ?? F ? ndS ? ?? ?
S
S1
S2
p? ?
2 p?
? ? p? ?
3 ?
3
Hence, we have verified the Gauss divergence theorem as
?? F ? ndS ? ??? divFdV ?
S
V
2 p?
3
Question 7:
The surface can be sketched as follow:
For given F , we can compute curlF as follow:
i
curl F ? ? ?x
y 2 ? x2
j
? ?y
z2 ? y2
k
? ?z
x2 ? z 2
??
? ??
??
?
?
?
?
?
? ? ? x2 ? z 2 ? ? ? z 2 ? y 2 ?? i ? ? ? x2 ? z 2 ? ? ? y 2 ? x2 ?? j ? ? ? z 2 ? y 2 ? ? ? y 2 ? x2 ?? k
?z
?z
?y
?
? ?y
? ? ?x
? ?x
?
? ? ?2 z ? i ? ? 2 x ? j ? ? ?2 y ? k
For f ? x, y, z ? ? x ? y ? z , we have ?f ? i ? j ? k , which implies n ?
1
?i ? j ? k ? .
3
Therefore,
1
? i ? j ? k ? dS
3
S
2
2
?
? x ? y ? z ? dS ? ?? pdS
??
3 S
3 S
?? curl F ? ndS ? ?? ?2 ? xi ? yj ? zk ? ?
S
2p
2p ? p 2?
?
? Area of triangle ?
?
3
2
4
3
3
? ? p3
On the other hand, we have
? F .d r ? ? F .d r ? ? F .d r ? ? F .d r
C
C1
C2
C3
For C1 , we have x ? y ? p and z ? 0 . Therefore,
? F .d r ? ? ? y
C1
2
? x 2 ? dx ? ? 02 ? y 2 ? dy ? 0
C1
p
??
?? p ? x ?
2
?
? x 2 dx ? ? p ? x ? ? ?dx ?
2
0
0
1 ?
3
? 2
= ? ? ? p ? x ? ? x3 ?
3 ?p
? 3
1
? ? p3
3
Similarly, we end up with
1
3
? 1
3
? F .d r ? ? F .d r ? ? 3 p
C2
C3
Therefore,
? F .d r ? ? F .d r ? ? F .d r ? ? F .d r ? ?? ? 3 p
C
C1
C2
C3
? ? 1 3? ? 1 3?
3
???? p ???? p ? ? ?p
? ? 3 ? ? 3 ?
Hence, we have verified the Stokes Theorem as
? F .d r ? ?? curl F .ndS ? ? p
C
S
3
Question 8:
?
?
We know that the Fourier transform of u0 ? t ? is F u0 ? t ? ?? ? ?
1
, and the Fourier
j?
transform of f ? t ? t0 ? is F ?? ? e? j?t0 , where F ?? ? is the Fourier transform of f ? t ? .
Now, let f ? t ? ? u0 ? t ? and t0 ? p , it follows that:
? The Fourier transform of u0 ? t ? p ? is F ?u0 ? t ? p ?? ?? ? ?
e j? p
j?
? The Fourier transform of u0 ? t ? p ? is F ?u0 ? t ? p ?? ?? ? ?
e? j? p
j?
? The Fourier transform of u0 ? t ? p ? ? u0 ? t ? p ? is
F ?u0 ? t ? p ?? ?? ? ? F ?u0 ? t ? p ?? ?? ? ?
?
e j? p e ? j? p e j? p ? e ? j? p 2sin ?? p ?
?
?
?
j?
j?
j?
?
?
?
Finally, the Fourier transform of tf ? t ? is F tf ? t ? ?? ? ? j
Fourier transform of f ? t ? . It implies that
dF ?? ?
where F ?? ? is the
d?
d ? 2sin ?? p ? ?
? F t ??u0 ? t ? p ? ? u0 ? t ? p ? ?? ?? ? ? j
?
?
d? ?
?
?
2? p cos ?? p ? ? 2sin ?? p ?
? j
2
?
?
?
Finally,
?A
?
A 2? p cos ?? p ? ? 2sin ?? p ?
? F ? t ??u0 ? t ? p ? ? u0 ? t ? p ? ?? ? ?? ? ? j
p
?2
?p
?
?
2 Aj
?? p cos ?? p ? ? sin ?? p ? ?
p? 2

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