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This preview shows page 1. to view the full 4 pages of the document. 1. [5 marks] Evaluate dydx
y
x
∫∫ +
1
0
1
4
3/1 1
1.
Since the integral cannot be evaluated directly, we have to reverse the order of
integration. The region of integration is , which can
also be described as . Hence we have
}10,1:),{( 3/1 = xyxyxD
}10,
3yy
0:),{( = xyxD
dydx
y
x+
1
0
1
4
3/1 1
1
∫∫ = dxdy
y
y
∫∫ +
1
00
4
3
1
1 = dy
y
y
+
1
0
4
3
1 = 1
0
4|)1ln(
4
1+y = 2ln
4
1
(b) [6 marks] Suppose that the mass density d of the body occupying the region V in the first octant
bounded by the cylinder , the plane
2
4yx = 1
=
+
zy , and the coordinate planes is given by
where f is continuous on V. Write an iterated integral which gives the mass m of the body. ),,( zyxfd =
Note that the plane intersects the xy-plane along the line y = 1. Projecting the
region on xy-plane we have
1=+ zy
m
∫∫∫
=
V
dVzyxf ),,( =
∫∫ ∫
−−
1
0
4
0
1
0
2
),,(
yy
dzdxdyzyxf
Projecting the region on the yz-plane we may write .
∫∫ ∫
=
1
0
1
0
4
0
2
),,(
zy
dxdydzzyxfm