Triple Integrals

TRIPLE INTEGRALS

The symbol _z1∫^z2 _y1∫^y2 _x1∫^x2 f(x, y, z) dxdydz denotes a triple integral .It means that f(x, y) is said to be integrated first with respect to x, between the limits x₁ and x₂ keeping y and z as constants temporarily the resulting function is then to be integrated with respect to y between the limits y₁ and y₂ .Keeping z as constant. the result thus obtained is finally integrated with respect to z between the limits z₁ and z₂. In this integral z₁ and z₂ are constants; y₁ and y₂ are either constants or functions of z alone and x₁, x₂ are either constants or functions of y and z.

 y₁ = f₁(z)

 y₂ = f₂(z)

 x₁ = g₁(y, z)

 x₂ = g₂(y, z)

The order in which the integrators are performed in the triple integral is illustrated in Figure 1.1

Note: In 2-D, xy - plane is divided into four quadrants. By the similar manner in 3-D, the region is divided into eight octants as follows.

Example 75. Evaluate ₀∫^log2₀∫^x₀∫^x+y e^x+y+z dx dy dz.

Solution:

Since inner limit depends on x, y, the function is integrated first w.r.t z. The middle integral depends on x, the function is integrated w.r.t y.

The correct form of the integral

Example 76. Evaluate ₀∫¹₀∫^1-x₀∫^1-x-y dzdydx.

Solution:

Example 77. Evaluate ∫∫∫ dxdydz / [a²‒x²‒y²‒z²] over the first octant of the sphere x² + y²+z² = a².

Solution:

In first octant all the lower limits in the triple integral are zero.

Given x²+ y²+ z² = a²         ………..(1)

z = √[a² ‒ x² ‒ y²]

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

Put z = 0 in (1), we get

 x² + y² = a²        ……..(2)

 y = √[ a² ‒ x²]

 y varies from y = 0 to y = √[a²‒x²]

Put y = 0 in (2)

(2) ⇒ x² = a²

⇒ x= a

  x varies from x = 0 to x = a

Example 78. Evaluate .

Solution:

Since inner limit depends on x, y, the function is integrated first w.r.t z. The middle integral depends on x, the function is integrated w.r.t y.

The correct form of the integral is

Example 79. Evaluate ∫∫∫ xyz dxdydz through the positive spherical octant for which x² + y² + z² = a². [or]

Evaluate ₀∫^a ₀∫^√[a2‒x2] ₀∫^{[√a2‒x2‒y2]}

Solution:

Since the region is positive octant, all the lower limits are zero.

x² + y² + z² = a²   ……….(1)

z² = a²‒x²‒y²

z = √[a²‒x²‒y²]

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

Put z = 0 in (1)

x² + y² = a²

y² = a² ‒ x²      ……..(2)

y = √[ a² ‒ x²]

 y varies from y = 0 to y = √[a²‒x²]

Put y = 0 in (2)

 x² = a² ⇒ x= a

 x varies from x = 0 to x = a

EXERCISE