Find the volume of sphere x2+y2+z2=a2
Answers
Answer:
First, we need to recall just how spherical coordinates are defined. The following sketch shows the relationship between the Cartesian and spherical coordinate systems.
This graph has a standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward. There is a point labeled $\left( x,y,z \right)=\left( r,\theta ,z \right)$ that appears to be in the 1st octant (i.e. x, y, and z are all positive). From this point a dashed line dropped straight down in the xy-plane (reaching it at a right angle) and the dashed line is labeled “z”. From the origin a new dashed line is drawn until it reaches the point where the “z” dashed line hits the xy-plane. The angle from the positive x-axis and the “r” dashed line is shown as $\theta$. In addition, there is a line from the origin up to the point that is labeled $\rho$ and the angle from the positive z-axis to this new line is shown at $\varphi $.
Here are the conversion formulas for spherical coordinates.
x
=
ρ
sin
φ
cos
θ
y
=
ρ
sin
φ
sin
θ
z
=
ρ
cos
φ
x
2
+
y
2
+
z
2
=
ρ
2
We also have the following restrictions on the coordinates.
ρ
≥
0
0
≤
φ
≤
π
For our integrals we are going to restrict
E
down to a spherical wedge. This will mean that we are going to take ranges for the variables as follows,
a
≤
ρ
≤
b
α
≤
θ
≤
β
δ
≤
φ
≤
γ
Here is a quick sketch of a spherical wedge in which the lower limit for both
ρ
and
φ
are zero for reference purposes. Most of the wedges we’ll be working with will fit into this pattern.
This is basically the sketch of an ice cream cone shaped solid. The sides are a cone that starts at the origin, is centered on the z-axis and opens up in the positive z direction. The cap is the portion of a sphere that fits on the top of the cone.
From this sketch we can see that
E
is nothing more than the intersection of a sphere and a cone and generally will represent a shape that is reminiscent of an ice cream cone.
In the next section we will show that
d
V
=
ρ
2
sin
φ
d
ρ
d
θ
d
φ
Therefore, the integral will become,
∭
E
f
(
x
,
y
,
z
)
d
V
=
∫
γ
δ
∫
β
α
∫
b
a
ρ
2
sin
φ
f
(
ρ
sin
φ
cos
θ
,
ρ
sin
φ
sin
θ
,
ρ
cos
φ
)
d
ρ
d
θ
d
φ
This looks bad but given that the limits are all constants the integrals here tend to not be too bad.
Example 1 Evaluate
∭
E
16
z
d
V
where
E
is the upper half of the sphere
x
2
+
y
2
+
z
2
=
1
.
Show Solution
Example 2 Evaluate
∭
E
z
x
d
V
where
E
is above
x
2
+
y
2
+
z
2
=
4
, inside the cone (pointing upward) that makes an angle of
π
3
with the negative
z
-axis and has
x
≤
0
.
Show Solution
Example 3 Convert
∫
3
0
∫
√
9
−
y
2
0
∫
√
18
−
x
2
−
y
2
√
x
2
+
y
2
x
2
+
y
2
+
z
2
d
z
d
x
d
y
into spherical coordinates.
Show Solution