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Concept Version 5
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Triple Integrals

For T⊆R3T \subseteq R^3T⊆R​3​​, the triple integral over TTT is written as ∭Tf(x,y,z)dxdydz\iiint_T f(x,y,z)\, dx\, dy\, dz∭​T​​f(x,y,z)dxdydz.

Learning Objective

  • Use triple integrals to integrate over three-dimensional regions


Key Points

    • By convention, the triple integral has three integral signs (and a double integral has two integral signs); this is a notational convention which is convenient when computing a multiple integral as an iterated integral.
    • If TTT is a domain that is normal with respect to the xyxyxy-plane and determined by the functions α(x,y)\alpha (x,y)α(x,y) and β(x,y)\beta(x,y)β(x,y), then ∭Tf(x,y,z) dxdydz=∬D∫α(x,y)β(x,y)f(x,y,z)dzdxdy\iiint_T f(x,y,z) \ dx\, dy\, dz = \iint_D \int_{\alpha (x,y)}^{\beta (x,y)} f(x,y,z) \, dz dx dy∭​T​​f(x,y,z) dxdydz=∬​D​​∫​α(x,y)​β(x,y)​​f(x,y,z)dzdxdy.
    • To integrate a function with spherical symmetry such as f(x,y,z)=x2+y2+z2f(x,y,z) = x^2 + y^2 + z^2f(x,y,z)=x​2​​+y​2​​+z​2​​, consider changing integration variable to spherical coordinates.

Term

  • spherical coordinate

    a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuth angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the zenith


Full Text

For T⊆R3T \subseteq R^3T⊆R​3​​, the triple integral over TTT is written as 

∭Tf(x,y,z)dxdydz\displaystyle{\iiint_T f(x,y,z)\, dx\, dy\, dz}∭​T​​f(x,y,z)dxdydz

Notice that, by convention, the triple integral has three integral signs (and a double integral has two integral signs); this is a notational convention which is convenient when computing a multiple integral as an iterated integral.

We have seen that double integrals can be evaluated over regions with a general shape. The extension of those formulae to triple integrals should be apparent. If TTT is a domain that is normal with respect to the xy-plane and determined by the functions α(x,y)\alpha (x,y)α(x,y) and β(x,y)\beta(x,y)β(x,y), then: 

∭Tf(x,y,z) dxdydz=∬D∫α(x,y)β(x,y)f(x,y,z)dzdxdy\displaystyle{\iiint_T f(x,y,z) \ dx\, dy\, dz = \iint_D \int_{\alpha (x,y)}^{\beta (x,y)} f(x,y,z) \, dz dx dy}∭​T​​f(x,y,z) dxdydz=∬​D​​∫​α(x,y)​β(x,y)​​f(x,y,z)dzdxdy

Graphical Representation of a Triple Integral

Example of domain in R3R^3R​3​​ that is normal with respect to the xyxyxy-plane.

Example 1

The volume of the parallelepiped of sides 4 by 6 by 5 may be obtained in two ways:

  • By calculating the double integral of the function f(x,y)=5f(x, y) = 5f(x,y)=5 over the region DDD in the xyxyxy-plane which is the base of the parallelepiped: ∬D5 dxdy\iint_D 5 \ dx\, dy∬​D​​5 dxdy
  • By calculating the triple integral of the constant function 1 over the parallelepiped itself: ∭parallelepiped1dxdydz\iiint_\mathrm{parallelepiped} 1 \, dx\, dy\, dz∭​parallelepiped​​1dxdydz

Example 2

Integrate f(x,y,z)=x2+y2+z2f(x,y,z) = x^2 + y^2 + z^2f(x,y,z)=x​2​​+y​2​​+z​2​​ over the domain D={x2+y2+z2≤16}D = \left \{ x^2 + y^2 + z^2 \le 16 \right \}D={x​2​​+y​2​​+z​2​​≤16}.

Looking at the domain, it seems convenient to adopt the passage in spherical coordinates; in fact, the intervals of the variables that delimit the new TTT region are obviously: 

(0≤ρ≤4, 0≤ϕ≤π, 0≤θ≤2π)(0 \le \rho \le 4, \ 0 \le \phi \le \pi, \ 0 \le \theta \le 2 \pi)(0≤ρ≤4, 0≤ϕ≤π, 0≤θ≤2π)

For the function, we get:

f(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)=ρ2f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) = \rho^2f(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)=ρ​2​​

Therefore:

∭D(x2+y2+z2)dxdydz=∭Tρ2 ρ2sinθdρdθdϕ=∫0πsinϕdϕ∫04ρ4dρ∫02πdθ=2π∫0πsinϕ[ρ55]04dϕ=2π[ρ55]04[−cosϕ]0π=4096π5\begin{aligned} \iiint_D (x^2 + y^2 +z^2) \, dx\, dy\, dz &= \iiint_T \rho^2 \ \rho^2 \sin \theta \, d\rho\, d\theta\, d\phi \\ &= \int_0^{\pi} \sin \phi \,d\phi \int_0^4 \rho^4 d \rho \int_0^{2 \pi} d\theta \\ &= 2 \pi \int_0^{\pi} \sin \phi \left[ \frac{\rho^5}{5} \right]_0^4 \, d \phi \\ &= 2 \pi \left[ \frac{\rho^5}{5} \right]_0^4 \left[- \cos \phi \right]_0^{\pi} = \frac{4096 \pi}{5} \end{aligned}​∭​D​​(x​2​​+y​2​​+z​2​​)dxdydz​​​​​​=∭​T​​ρ​2​​ ρ​2​​sinθdρdθdϕ​=∫​0​π​​sinϕdϕ∫​0​4​​ρ​4​​dρ∫​0​2π​​dθ​=2π∫​0​π​​sinϕ[​5​​ρ​5​​​​]​0​4​​dϕ​=2π[​5​​ρ​5​​​​]​0​4​​[−cosϕ]​0​π​​=​5​​4096π​​​​

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