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,

3≤≤ yy

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

Other answers possible.

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