Triple Integrals: Volume of Solids

 

VOLUME OF SOLIDS

 

Definition:

The volume V of three dimensional region is given by V = ∫∫∫_V dzdydx. If the region is bounded by

z = f₁(x,y), z = f₂(x, y), y = Ø₁(x), y = Ø₂(x) and x = a, x = b then

The order of integration can be changed with a suitable change in the limits of integration.

In cylindrical polar coordinate system, the volume of a region of space V is given by

In spherical polar coordinate system, the volume of a region of space V is given by

 

Example 87. Find the volume of the region bounded by x = 0, y = 0, z = 0, x/a + y/b + z/c = 1.

Solution:

To plot the region:

Given region is bounded by the planes x = 0, y = 0, z = 0 and x/a + y/b + z/c = 1.

x/a + y/b + z/c = 1 is plane which intersect the coordinate axes at (a, 0,0), (0, b, 0), (0,0, c)

To evaluate the volume:

Volume V = ∫∫∫ dz dy dx

To find the limits:

Given x/a + y/b + z/c = 1 .......(1)

 ⇒ z = c ( 1 – x/a – y/b)

 z varies from z = 0 to z = c(1 – x/a – y/b)

Put z = 0(1), the projection of the solid in the xy - plane is the region bounded by x = 0, y = 0 and x/a + y/b = 1  .......(2)

y = b(1 ‒ x/a)

 y varies from y = 0 to y = b (1 ‒ x/a)

Put y = 0 in (2), we get x/a = 1

x = a

x varies from x = 0 to x = a

Volume V = ∫∫∫_V dz dy dx

Put, t = 1‒x/a ; don’t find dt because 1‒x/a is a constant with respect to y

Volume = abc/6 cubic units.

 

Example 88. Find the volume of the sphere of radius 'a'.

Solution:

To plot the region:

Equation of a sphere of radius 'a' is x² + y² + z² = a²       …….(1)

To evaluate the volume:

Volume V = 8 × Volume of the solid lies in the first octant

= 8 × ∫∫∫_V dz dy dx

To find the limits:

 (1) ⇒ z² = a² ‒ x² ‒ y²

 z = √[a² ‒ x² ‒ y²] (Since the region is first octant)

 z varies from z = 0 to z = √[a² ‒ x² ‒ y²]

Put z = 0 in (1), the projection of the solid in the xy‒plane is the circle

x² + y² = a²    ... (2)

⇒ y = √[a² ‒ x²]

 y varies from y = 0 to y = √[a²‒ x²]

Put y = 0 in (2), we get x² = a² ⇒ x= a

x varies from x = 0 to x = a

V = 4/3 πа³ cubic units.

 

Note: We can find the volume of the sphere using spherical polar coordinates system

 

Example 89. Find the volume of the sphere of radius 'a' using spherical polar coordinates system

Solution:

To plot the region:

Equation of a sphere of radius 'a' is x² + y² + z² = a²         ….. (1)

To evaluate the volume:

Volume V = 8 × Volume of the solid lies in the first octant

= 8 × ∫∫∫_V dz dy dx

To convert from Cartesian to spherical polar coordinates system, we have the following transformation

x = r sinθ cosϕ

y = r sinθ sinϕ

z = r cosθ

dxdydz = r² sinθdrdθdϕ

Also x² + y² + z² = r²

The region is the volume of the first octants sphere x² + y² + z² = a².

Therefore the limits are

r = 0; r = a

θ = 0; θ = π/2

ϕ = 0; ϕ = π/2

Volume of sphere

 

Example 90. Find the volume of the ellipsoid 

Solution:

To plot the region:

 is an ellipsoid.

To evaluate the volume:

Volume V = 8 × Volume of the solid lies in the first octant

= 8 ∫∫∫_V dzdydx

To find the limits:

Put z = 0 in (1), the projection of the solid in the xy - plane is the ellipse

 

Example 91. Find the volume common to the cylinders x² + y² = a² and x² + z² = a².

Solution:

To plot the region:

 x² + z² = a² and x² + y² = a² are cylinders which are symmetrical about y - axis and z ‒ axis respectively.

The section of the cylinder x²+z² = a² is the circle x² + y² = a² in the xy - plane. In the figure only one‒eight (in particular octant) of the required volume is shown.

To evaluate the volume:

Volume V = 8 × Volume of the solid lies in the first octant

= 8 ∫∫∫_V dzdydx

To find the limits:

We have x² + z² = a²

 z² = a² ‒ x² ⇒ z = √[a² = x²]

..z varies from

z = 0 to z = √[a² ‒ x²]

The projection of the solid in the xy planeis x² + y² = a²        ... (1)

which is a circle in the xy ‒ plane.

⇒ y² = a² ‒ x² ⇒ y = √[a² ‒ x²]

. y varies from y = 0 to y=√[a² ‒ x²]

Put y = 0 in (1)

x² = a² ⇒ x = a

.x varies from x = 0 to x = a

 

Example 92. Find the volume bounded by the cylinder x² + y² = 4, y+z = 4, z = 0.

Solution:

To plot the region:

x² + y² = 4   ... (1) is a cylinder which are symmetrical about z-axis.

y + z = 4  ... (2) is a plane intersecting y and z axes.

z = 0 is xoy plane. The common volume is given in the following figure

To evaluate the volume:

Volume V = ∫∫∫_V dz dy dx

Since the region is cylindrical, we use cylindrical polar coordinates system.

We have the following transformation

x = r cosθ, y = r sinθ, z = z

dx dy dz = r dr dθ dz.

Also x² + y² = r².

To evaluate the volume:

Volume V = ∫∫∫_V dz dy dx

To find the limits:

Given z = 0 and y + z = 4

z = 4‒y = 4‒r sinθ

 z varies from z = 0 to z = 4 ‒ 4 ‒ rsinθ

x² + y² = 4

 r² = 4 ⇒ r=2

 r varies from r = 0 to r=2

In a cylinder, 0 ≤ θ ≤ 2π

V = 16π cubic units.

 

Example 93. Find the volume of the paraboloid of revolution x² + y² = 4z cut off by the plane z = 4.

Solution:

To plot the region:

Given x² + y² = 4z          ……..(1) is a paraboloid which is cut by the plane z = 4 shaded region. Put z = 4 in (1), we get x² + y² = 16

The section of the paraboloid by the plane z = 4 is the circle x² + y² = 16        ... (2)

The projection of the given solid on the xy‒plane is the circle x²+ y² = 16.

Since the region is cylindrical, we use cylindrical polar coordinates system.

We have the following transformation

x = r cosθ, y = r sinθ, z = z

dx dy dz = r dr dθ dz.

Also x²+y² = r²,

To evaluate the volume:

Volume V = ∫∫∫_V dz dy dx

To find the limits:

Now from (1),

z = x²+ y² / 4

z = r²/4

 z varies from z = r²/4 to z = 4

(2) ⇒ r² = 16 ⇒ r = 4

 r varies from r = 0 to y=4

In a cylinder, 0 ≤ θ ≤ 2π.

V = 32π cubic units.

 

Example 94. Find the volume of the portion of the cylinder x²+ y² = 1 intercepted between the plane z = 0 and the paraboloid x² + y² = 4‒z.

Solution:

To plot the region:

The region is the cylinder x² + y² = 1 intercepted between the plane z = 0 and the paraboloid x² + y² = 4‒z

Since the region is cylindrical, we use cylindrical polar coordinates system.

We have the following transformation

 x = r cosθ, y = r sinθ, z= z

dx dy dz = r dr dθ dz.

Also x²+ y²= r².

x² + y² = 1

r² = 1.

⇒ 0 ≤ r ≤ 1.

In a cylinder, 0 ≤ θ ≤ 2π.

Given z = 4 ‒ x² ‒ y²

= 4‒(x² + y²)

= 4‒r²

 z varies from z = 0 to z = 4 ‒ r²

To evaluate the volume:

 

Example 95. Find the volume of a solid bounded by the spherical surface x² + y² + z² = 4a² and the cylinder x² + y² ‒ 2ay = 0.

Solution:

Cylindrical co‒ordinates

x = r cosθ

y = r sinθ

z = z

x² + y² = r²

dz dy dx = r dz dr dθ

The equation of the sphere and the cylinder

x² + y² + z² = 4a²

r² + z² = 4a²

z² = 4a² ‒ r²

z = √[4a² ‒ r²]

and the cylinder x² + y² ‒ 2ay = 0

x² + y² = 2ay

 r² = 2ar sinθ

 r = 2a sinθ

Therefore the limits are

z=0; z = √[4a² ‒ r²]

r = 0; r = 2a sinθ

θ = 0; θ = π/2

The volume inside the cylinder bounded by the sphere is twice the volume shown shaded in the Figure for which z varies from 0 to √[(a² ‒ r²)],

Hence, the required volume is given by

Volume = ∫∫∫ dz dy dx

 

Example 96. Find the volume bounded by the paraboloid x² + y² = az, the cylinder x² + y² = 2ay and the plane z = 0.

Solution:

The required volume is founded by integrating z = x²+y² / a, over the circle x²+y² = 2ay

Changing to polar coordinates in the xy‒plane, we have x = r cosθ, y = r sinθ

x² + y² = r²

dx dy = rdr dθ

We have the circle.

x² + y² = 2ay

r² = 2ar sinθ

r = 2a sinθ

z = x²+ y² / a

⇒ z = r²/a

and the polar equation of the circle is r = 2a sinθ.

To cover this circle, r varies from 0 to 2a and θ varies from θ = 0 to π.

Hence the required volume is

 

 

EXERCISE

 

82. Find the volume of the region bounded by x = 0, y = 0, z=0, x + y + z = 1

Ans: 1/6

 

83. Find the volume of the region bounded by x = 0, y = 0, z = 0, x + 2y + 3z = 6.

Ans: 6 cubic units

 

84. Find the volume of the sphere of radius 'a' which lies in the first octant.

Ans: πa³/6

 

85. Find the volume of the spherical x² + y² + z² = 3² by using triple integration.

Ans: 36π

 

86. Find the volume of the ellipsoid x²/a² + y²/b² +  z²/c² = 1 which lies in the first octant.

Ans: πabc / 6

 

87. Find the volume of the ellipsoid x²/4 + y²/9 + z²/16 = 1.

Ans: 32π

 

88. Find the volume bounded by the cylinder x² + y² = 4 and the planes y + z = 3 and z = 0.

Ans:12π

 

89. Find the volume of the cylinder x² + y² = a², z = 0 & z= h.

Ans: πa²h

 

90. Find the volume of the region bounded by z = x² + y², z=0, x = ‒a, x = a and y = ‒a and y = a.

Ans: 8a⁴ /3 cubic units.

 

91. Find the volume of the paraboloid of revolution x² + y² = z cut off by the plane z = 4.

Ans: 2π cubic units

 

92. Find the volume of the region bounded by the surfaces y² = 4ax and x² = 4ay and the plane z = 0 z = 3.

Ans: Volume = 16 a² cubic units

 

93. Calculate the volume of the solid bounded by the planes x = 0, y = 0, 2x + 3y+4z = 12 and z = 0.

Ans: 12

 

94. Find the volume of the portion of the sphere x² + y² + z² = a² lying inside the cylinder x² + y² = ay.

Ans: 2a³/9 [3π ‒ 4]