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F18XC1 Solutions 8: Applications of Multiple Integrals

8.1.

x

r= a

y

For C the range is 0 < 2 and the r rangeis 0< r < a. Hence

C

x2 +y2 dA=

2

0

a0

r r dr

d=

2

0

a3

3 d= 2a3/3 .

8.2.

(a) A sketch of the region is as follows:

(3, 0)x

0

r= 3

(0, 3)

y

The range is 0< < /2 and the r range is 0 < r

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Set u = 1 + r2 so du = 2r dr and

1 +r2 r dr = 12

u

1

2du = 12 2

3u

3

2 = 13

(1 +r2)3

2 .Then

1/

2

0

1x2

x

1 +x2 +y2 dy dx=

/2/4

1

3(1 +r2)

3

2

10

d=

41

3(

81) = 12

(2

21) .

8.3.

(i) r= 1

x

y

For D the range is 0< < and the r range is0< r

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8.4.

Here is a sketch of the situation:

The paraboloid crosses the x - y plane where z= 0, i.e. on the circle x2 +y2 = r2 = 4.

Hence the region of integration is Cgiven by 0 <

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8.5.

The region is as shown on the right. We wish to calculatethe volume, which is given by

T(x2 + 2y2) dA. For T,

the x range is 0 < x < 1 and the y range (given x) is0< y 0, y > 0, with2x+ y < 2) and below the given plane, i.e. thevolume is

T(22xy) dA. ForTthexrange is 0 < x

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8.8.

x

y= x

(0, 0) (2, 0)

(1, 1) Ty

y= 2 x

The triangle, say T, lies in x >0 and is boundedby the lines x = y and x+ y = 2. Hence theyrange is 0 < y < 1 while, given y, the x range isy < x