MA 22S1
Assignment 9
Due never
Id:
22S1-f2015-9.m4,v 1.1 2016/01/10 00:47:22 john Exp john
1. Convert the triple integral
Z
√
1+ 2
√
1− 2
Z
√
1+ 1+2x−x2
√
1− 1+2x−x2
Z
√
−
2x+2y−x2 +y 2
√
(x2 + y 2 + z 2 ) dz dy dx
2x+2y−x2 +y 2
to cylindrical and spherical coordinates. Do not evaluate.
Solution: We need to figure out three things: the limits of integration,
the integrand, and the Jacobian factors.
The Jacobian factors are easy, since they depend only on the choice
of coordinate system. They are just r for cylindrical coordinates and
ρ2 sin φ for spherical coordinates.
The integrand is relatively straightforward:
x2 + y 2 + z 2 = r 2 + z 2 = ρ2 .
The hard part is the limits. The region of integration is described by
the inequalities
√
√
1 − 2 ≤ x ≤ 1 + 2,
√
√
1 − 1 + 2x − x2 ≤ y ≤ 1 + 1 + 2x − x2 ,
q
− 2x + 2y − x2 − y 2 ≤ z ≤
q
2x + 2y − x2 − y 2.
We can replace the last pair by the single inequality
z 2 ≤ 2x + 2y − x2 − y 2,
1
Id:
22S1-f2015-9.m4,v 1.1 2016/01/10 00:47:22 john Exp john 2
or, equivalently,
x2 + y 2 + z 2 − 2x − 2y ≤ 0
or
(x − 1)2 + (y − 1)2 + z 2 ≤ 2.
The other pairs of inequalities follow from this one. For example, (y −
1)2 ≥ 0 and z 2 ≥ 0 so (x − 1)2 ≤ 2 and hence
√
√
1 − 2 ≤ x ≤ 1 + 2.
We conclude that the region is described by the single inequality
(x − 1)2 + (y − 1)2 + z 2 ≤ 2.
√
This is just the ball of radius 2 about the point (1, 1, 0).
We now need to convert this to cylindrical and spherical coordinates.
For this it’s easier to use the equivalent form
x2 + y 2 + z 2 − 2x − 2y ≤ 0.
In cylindrical coordinates this is
r 2 + z 2 − 2r cos θ − 2r sin θ ≤ 0
and in spherical coordinates it is
ρ2 − 2ρ sin φ cos θ − 2ρ sin φ sin θ ≤ 0.
To get the limits in cylindrical coordinates we need to figure out the
possible values of θ, the possible values of r for given θ, and then the
possible values of z for given r and θ. Since r ≥ 0 and z 2 ≥ 0 we have
r ≤ 2 cos θ + 2 sin θ and hence 2 cos θ + 2 sin θ ≥ 0, which is possible
only if
−π/4 ≤ θ ≤ 3π/4.
The allowed values of r are then
0 ≤ r ≤ 2 cos θ + 2 sin θ.
Finally, the allowed values of z are
√
√
− 2r cos θ + 2r sin θ − r 2 ≤ z ≤ 2r cos θ + 2r sin θ − r 2 .
Id:
22S1-f2015-9.m4,v 1.1 2016/01/10 00:47:22 john Exp john 3
The integral in cylindrical coordinates is therefore
Z
3π/4
−π/4
Z
2 cos θ+2 sin θ
0
Z
√
2r cos θ+2r sin θ−r 2
(r
√
− 2r cos θ+2r sin θ−r 2
2
+ z 2 ) r dz dr dθ.
In spherical coordinates the allowed values of θ are the same:
−π/4 ≤ θ ≤ 3π/4.
The equation ρ2 − 2ρ sin φ cos θ − 2ρ sin φ sin θ ≤ 0 then doesn’t impose
any restrictions on φ other than the usual requirement that
0 ≤ φ ≤ π.
For given θ and φ the range of possible values of ρ is then
0 ≤ ρ ≤ 2 sin φ cos θ + 2 sin φ sin θ.
The integral is then
Z
3π/4
−π/4
Z
0
π
Z
2 sin φ cos θ+2 sin φ sin θ
ρ4 sin φ dρ dφ dθ.
0
Note that the choice of order of integration in each case is arbitrary.
If you order the variables differently you will get different limits of
integration, although the integrand should be unchanged.