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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
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