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Triple integrals
Spherical regions and triple integrals in spherical coordinates suggested problems – solutions
P1: Describe and sketch the region whose volume is given by
2 4
0 0
/2 2
/6sin d dd
π π
πρ φ ρ φ θ∫ ∫ ∫
Then evaluate the integral to find the volume.
The region is
, , ) | 0 4, , 2 }(6 2
{ 0π πρ φ θ ρ φ θ π≤ ≤ ≤ ≤ ≤ ≤
so it looks like a wedge of a spherical solid – the ρ values are giving us a sphere of radius 4 and the θ
values are giving us a full circle, but the tilt runs from 6 2π πφ≤ ≤ relative to the z axis (note a tilt of
2π
from z just puts you in the xy plane). Having tried and discarded a few variations, I’ve come to the
conclusion that it’s easier to just hand sketch these things than try to coax any plotting software into producing a reasonable picture. That said, it’s also a pain to hand sketch – the problem isn’t perspective, but transparency; looking at this shape face on, you’ll just see the outer rounded shell, but the inside is cone shaped. Try doing 3d sketches accompanied by 2d cross sections to get the point across.
Evaluating the integral:
Inner:
42 3
0
3 3
4
0
1 sisin
64 sin
n31 s 4
3in 0
3
dρ φ ρ φ ρ
φ φ
⎡ ⎤⎣ ⎦
⎡ ⎤ =⎦
=
= −⎣
∫
Middle:
[ ]/2 /2
/6/6
64 cos364 64 3 32 3c
6
o
4 sin3
s cos 03 2 6 3 2 3
dπ π
ππφ φ φ
π π
= −
⎡ ⎤⎡ ⎤− − = − − =⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦=
∫
Outer:
[ ]2 2
00
32 3 32 3 3 1163 3
643
dπ πθ θ π ≈==∫
The volume is approximately 116 cubic units. Note that the 2 sinρ φ is picked up as part of the integral
when working in spherical regions; the function being integrated is simply , , 1( )f ρ θ φ = , and so that
integral is in fact the volume.
Maple check:
P2: Convert the integral from rectangular to spherical coordinates and evaluate.
2 2 2
2
4
2
42 2
4 2
x x y
xx dy ddz x
− − −
− −
+
−∫ ∫ ∫
There’s a trick to this one, and without it, the integral is difficult to impossible. Note that z lies between
2z = and 2 22 4 yz x− −= + . The “2+” part would make this miserable in spherical, except for one
thing…the 2 24 xz y− −= is just the top half of the sphere 2 2 2 4yx z+ + = , and the x and y
bounds are telling us it’s the full hemisphere, and not just a wedge of it. So what that “2+” is doing is just boosting the sphere up 2 units…and the lower bound of 2z = is boosting the plane up with it. So we still have a nice neat hemisphere, floating 2 units above the xy plane:
Now, look at the integrand. If it depended on z (so the height of the region mattered), we’d be in trouble, but it doesn’t – it’s solely a function of x . So it’s not going to matter if we drop that region back
down 2 units and set it back in the plane – the integral 2 2 2
2
4
2
42 2
4 2
x x y
xx dy ddz x
− − −
− −
+
−∫ ∫ ∫ is equivalent to
the integral 2 2 2
2
2
2
4 4
4 0
x x y
xx dy dxdz
− − −
− −−∫ ∫ ∫ . And that’s easy to set up in spherical.
The hemisphere of radius 2 is described by
, , ) | 0 2,0 ,0 2 }2
{( πρ φ θ ρ φ θ π≤ ≤ ≤ ≤ ≤ ≤
The integrand ( , , )f x y z x= becomes , , ) sin cos(f ρ θ φ ρ φ θ= .
And dz dy dx becomes 2 sin d d dρ φ ρ φ θ .
So we now have
( )/2 /22 3 2
0 0
2 2 2 2
0 0 0 0s sinin cos ci os n sd d d dd d
π π π πρ φ θ ρ φ ρ φ θ ρ φ θ ρ φ θ=∫ ∫ ∫ ∫ ∫ ∫
Inner:
23 2 2
0
2
2 4 4
4
2
0sin
4
1cos sin cos41 sin cos 2 0 sin cos4
dρ φ θ ρ φ θ ρ
φ θ φ θ
⎡ ⎤⎣ ⎦
⎡ ⎤= =⎣ ⎦
=
−
∫
Middle:
/2/2 2
00
sin cossin cos 4cos2
sin cos 0 sin 0cos02 2 24cos 4cos cos2 2 4
4 dπ
π φ φ φφ θ φ θ
π π ππθ θ θ
−⎡ ⎤= ⎢ ⎥⎣ ⎦
⎡ ⎤−⎢ ⎥− ⎡ ⎤− = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦
=
∫
Outer:
[ ]2 2
00cos sin 0d
π πθ θ θ= =∫
All that to get 0! We can do a Maple check (and I will), but there’s also a geometry check – look at the symmetry. The region over which we’re integrating covers the top four octants in 3D, and is symmetric . In two of those four octants, x is positive (the two that lie above the first and third quadrants), and in the other two (above the second and fourth quadrants), x is negative. Since the thing we are integrating is in fact x , the positive and negative contributions balance out.
Maple check:
And Mapling the original:
P3: Use spherical coordinates to find the volume of the solid between the spheres 2 2 2 4yx z+ + = and 2 2 2 9yx z+ + = , and inside the cone 2 2 2z x y= + .
Again, this region is probably easier to just hand sketch than it is to try and coax something into plotting
it. The spheres are obvious, and you need to recognize 2 2 2z x y= + as the double cone (solved for z it’s both plus and minus) with the sides at a 45°angle. I’m not even going to try to draw the double cone with the 3d perspective lines, only the part that lies above the xy plane. This is another one that’s difficult to show the transparency – the outer shell that you’d see would be a wedge of a sphere of
radius 3 ‐ it’s just that instead of a solid sphere, it’s a spherical shell with an inner radius of 2 .
This is one where cross sections are helpful:
We’ll make use of symmetry and just do the top half and double it. To stay in the cone, we need to run
the tilt from 0 to 4π, and the inner and outer radii of the spheres are 2ρ = and 3ρ = , so the region is
described by
, , ) | 2 3,0 ,0 2 }4
{( πρ φ θ ρ φ θ π≤ ≤ ≤ ≤ ≤ ≤
The half volume is given by
/4 22
0 20
3sin
DddV d d
π πρ φ ρ φ θ=∫ ∫ ∫ ∫
Inner:
3 32 3
3 3
22sin
193 sin3
1 sin31 sin 23
dρ φ ρ φ ρ
φ φ
⎡ ⎤⎣ ⎦
⎡ ⎤= =⎣ ⎦
=
−
∫
Middle:
[ ]/4 /4
00
19 cos3
19 19 2 19(2 2)cos cos0 13 4 3
19 sin3
2 6
dπ πφ φ φ
π
= −
⎡ ⎤ −⎡ ⎤− − = − − =⎢ ⎥⎦
= ⎢ ⎥⎣ ⎦ ⎣
∫
Outer:
[ ]2 2
00
19(2 2) 19(2 2) 19 (2 2)6 6 3
dπ π πθ θ− − −
= =∫
Total volume is
38 (2 2) 23.33
π −≈ .
Maple check:
P4: Same region as P3 but outside the cone…
Again, showing the top only – so I’m running the tilt from 4πto
2π. The full double cone wedge (shown
in the cross section) would run from 4πto
34π
, or you could set up like the previous one, do just the top
half, and double it.
This one has the same integration as P3, except the φ values run 4πto
2πnow (again, we’ll make use of
symmetry and just do the top half and double it). The region is described by
, , ) | 2 3, , 2 }(4 2
{ 0π πρ φ θ ρ φ θ π≤ ≤ ≤ ≤ ≤ ≤
The half volume is given by
/2 2
/ 24
2 3
0sin
DdV d d d
π π
πρ φ ρ φ θ=∫ ∫ ∫ ∫
Inner:
3 32 3
3 3
22sin
193 sin3
1 sin31 sin 23
dρ φ ρ φ ρ
φ φ
⎡ ⎤⎣ ⎦
⎡ ⎤= =⎣ ⎦
=
−
∫
Middle:
[ ]/2 /2
/4/4
19 cos3
19 19 2 19 2c
1
o
9 sin3
s cos 03 2 4 3 2 6
dπ π
ππφ φ φ
π π
= −
⎡ ⎤⎡ ⎤− − = − − =⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦=
∫
Outer:
[ ]2 2
00
19 2 19 2 19 26 6 3
dπ π πθ θ= =∫
Total volume is
38 2 56.33π
≈ .
I won’t Maple check again, since same integrals, but how about a geometry check? The total of the two volumes in P3 and P4 should be equal to the volume of the spherical shell.
Volume of sphere of radius 3: 34 1083 3
(3) ππ = .
Volume of sphere of radius 2: 34 323
)3
(2 ππ = .
Difference: 108 32 76
3 3 3π π π− = .
Total from P3 and P4: 38 (2 2) 38 2 38 (2 2 2) 38 (2) 76
3 3 3 3 3π π π π π− − +
+ = = =
Checks.