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   Calculation of Volumes Using Triple Integrals
The volume of a solid U in Cartesian coordinates xyz is given by
volume of a solid in Cartesian coordinates
In cylindrical coordinates, the volume of a solid is defined by the formula
volume of a solid in cylindrical coordinates
In spherical coordinates, the volume of a solid is expressed as
volume of a solid in spherical coordinates
   Example 1
Find the volume of the cone of height H and base radius R (Figure 2).

Solution.
volume of the cone
Fig.1
The cone is bounded by the surface and the plane z = H (Figure 1). Its volume in Cartesian coordinates is expressed by the formula
     
Calculate this integral in cylindrical coordinates that range within the limits:
     
As a result, we obtain (do not forget to include the Jacobian ρ):
     
Then the volume of the cone is
     
   Example 2
Find the volume of the ball x2 + y2 + z2R2.

Solution.
We calculate the volume of the part of the ball lying in the first octant (x ≥ 0, y ≥ 0, z ≥ 0), and then multiply the result by 8. This yields:
     
As a result, we get the well-known expression for the volume of the ball of radius R.


   Example 3
Find the volume of the tetrahedron bounded by the planes passing through the points A (1;0;0), B (0;2;0), C (0;0;3) and the coordinate planes Oxy, Oxz, Oyz (Figure 2).
volume of the tetrahedron
Fig.2
Fig.3

Solution.
The equation of the straight line AB in the xy-plane (Figure 3) is written as y = 2 − 2x. The variable x ranges here in the interval 0 ≤ x ≤ 1, and the variable y ranges in the interval 0 ≤ y ≤ 2 − 2x.

Write the equation of the plane ABC in segment form. Since the plane ABC cuts the line segments 1, 2, and 3, respectively, on the x-, y-, and z-axis, then its equation can be written as
     
In general case the equation of the plane ABC is written as
     
Hence, the limits of integration over the variable z range in the interval from z = 0 to . Now we can calculate the volume of the tetrahedron:
     
   Example 4
Find the volume of the tetrahedron bounded by the planes x + y + z = 5, x = 0, y = 0, z = 0 (Figure 4).

Solution.
The equation of the plane x + y + z = 5 can be rewritten in the form
     
By setting z = 0, we get
     
Fig.4
Fig.5
Hence, the region of integration D in the xy-plane is bounded by the straight line y = 5 − x as shown in Figure 5.

Representing the triple integral as an iterated integral, we can find the volume of the tetrahedron:
     
   Example 5
Find the volume of the solid formed by two paraboloids:
     
Fig.6
Fig.7

Solution.
Investigate intersection of the two paraboloids (Figure 6). Since ρ2 = x2 + y2, the equations of the paraboloids can be written as
     
By setting z1 = z2 for the intersection curve, we obtain
     
For this value of ρ (Figure 7), the coordinate z is
     
The volume of the solid is expressed through the triple integral as
     
This integral in cylindrical coordinates becomes
     
   Example 6
Calculate the volume of the ellipsoid
     

Solution.
It is easier to calculate the volume of the ellipsoid using generalized spherical coordinates. Let
     
Since the absolute value of the Jacobian for transformation of Cartesian coordinates into generalized spherical coordinates is
     
hence,
     
The volume of the ellipsoid is expressed through the triple integral:
     
By symmetry, we can find the volume of 1/8 part of the ellipsoid lying in the first octant (x ≥ 0, y ≥ 0, z ≥ 0), and then multiply the result by 8. The generalized spherical coordinates will range within the limits:
     
Then the volume of the ellipsoid is
     
   Example 7
Find the volume of the solid bounded by the sphere x2 + y2 + z2 = 6 and the paraboloid x2 + y2 = z.

Solution.
We first determine the curve of intersection of these surfaces. Substituting the equation of the paraboloid into the equation of the sphere, we find:
     
The second root z2 = −3 corresponds to intersection of the sphere with the lower shell of the paraboloid. So we do not consider this case. Thus, intersection of the solids happens at z = 2. Obviously, the projection of the region of integration on the xy-plane is the circle (Figure 8) defined by the equation x2 + y2 = 2.
Fig.8
Fig.9
The region of integration is bounded from above by the spherical surface, and from below by the paraboloid (Figure 9).
The volume of the solid region is expressed by the integral
     
It is convenient to convert the integral to cylindrical coordinates:
     
where ρ2 = x2 + x2 and the integral includes the Jacobian ρ. As a result, we have:
     
We change the variable: ρ2 = t. Here t = 0 when ρ = 0, and, respectively, t = 2 when ρ = √2.

Now we can calculate the volume of the solid:
     
   Example 8
Calculate the volume of the solid bounded by the paraboloid  z = 2 − x2y2 and the conic surface
.

Solution.
First we investigate intersection of the two surfaces. By equating the coordinates z, we get the following equation:
     
Let x2 + y2 = t2. Then
     
Only the root t = 1 has the sense in the context of the given problem, i.e.
     
Thus, both the surfaces intersect at z = 1, and the intersection is a circle (Figure 10).
Fig.10
Fig.11
The region of integration is bounded from above by the paraboloid, and from below by the cone (Figure 11).
To calculate the volume of the solid we use cylindrical coordinates:
     
As a result, we find:
     

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