Multiple Integrals

Multiple Integrals:

  1. Definition:

    • Multiple integrals extend the concept of single integrals to functions of multiple variables. They involve integrating functions over regions in higher-dimensional spaces.
  2. Types of Multiple Integrals:

    a. Double Integrals:

    • Involves integrating a function of two variables over a region in the plane.
    • Denoted as Rf(x,y)dAwhere R is the region in the xy-plane and dArepresents the differential area element.
    • Computational methods include iterated integrals (Fubini's theorem) and polar, rectangular, or other coordinate transformations.

    b. Triple Integrals:

    • Extends the concept to integrate a function of three variables over a region in 3D space.
    • Denoted as Vf(x,y,z)dV where V represents a volume in 3D space and dV is the differential volume element.
    • Methods involve iterated integrals in various coordinate systems (rectangular, cylindrical, spherical) and applications in physics, engineering, and geometry.
  3. Iterated Integrals:

    a. Double Integrals as Iterated Integrals:

    • Expresses a double integral as an iterated integral, computing one integral with respect to one variable and then the other.
    • For a function f(x,y), the double integral Rf(x,y)dA can be expressed as abcdf(x,y)dydxor vice versa.

    b. Triple Integrals as Iterated Integrals:

    • Similar to double integrals, expresses a triple integral as iterated integrals in various orders based on the application and convenience.
  4. Applications:

    • Physics: Used in calculating mass, moments, center of mass, and physical quantities in 3D space (like electric charge distribution).

    • Engineering: Applied in solving problems related to fluid dynamics, heat transfer, and material science involving volumes and fluxes in three dimensions.

    • Probability and Statistics: Utilized to calculate joint probabilities, expected values, and moments in multiple dimensions.

  5. Properties of Multiple Integrals:

    • Linearity, additivity, and properties of single integrals extend to multiple integrals.
    • Changing the order of integration often simplifies calculations, especially when dealing with certain regions or coordinate systems.

Example: Double Integral Calculation

Consider the function f(x,y)=x2+y over the region R defined by 0x2 and 1y3. Let's find the value of the double integral Rf(x,y)dA.

  1. Objective:

    • Calculate the double integral of f(x,y)=x2+y over the region R defined by 0x2 and 1y3.
  2. Solution Steps:

    • Step 1: Setting Up the Integral:

      • Express the double integral Rf(x,y)dA as abcdf(x,y)dydx over the region R.
      • For f(x,y)=x2+y, the integral becomes: R(x2+y)dydx over 0x2 and 1y3
    • Step 2: Evaluate the Inner Integral:

      • Integrate with respect to y first: 13(x2+y)dy =[x2y+y22]13=(3x2+92)(x2+12) =2x2+4
    • Step 3: Evaluate the Outer Integral:

      • Integrate the result from the inner integral with respect to x over the given interval: 02(2x2+4)dx =[2x33+4x]02=(163+8)(0) =403
  3. Result:

    • The value of the double integral R(x2+y)dydx over the region R defined by 0x2 and 1y3 is 403.

Example: Triple Integral Calculation

Consider the function f(x,y,z)=x+y+z over the region V defined by 0x1, 0y2, and 0z3. Let's find the value of the triple integral Vf(x,y,z)dV.

  1. Objective:

    • Calculate the triple integral of f(x,y,z)=x+y+z over the region V defined by 0x1, 0y2, and 0z3.
  2. Solution Steps:

    • Step 1: Setting Up the Integral:

      • Express the triple integral Vf(x,y,z)dV as abcdpqf(x,y,z)dzdydx over the region V.
      • For f(x,y,z)=x+y+z and the given limits, the integral becomes: V(x+y+z)dzdydx over 0x1,0y2,0z3
    • Step 2: Evaluate the Inner Integral:

      • Integrate with respect to z first: 03(x+y+z)dz =[xz+yz+z22]03=(3x+3y+92)(0) =3x+3y+92
    • Step 3: Evaluate the Intermediate Integral:

      • Integrate the result from the inner integral with respect to y over the given interval: 02(3x+3y+92)dy =[3xy+3y22+9y2]02=(6x+6+9)(0) =6x+15
    • Step 4: Evaluate the Outer Integral:

      • Integrate the result from the intermediate integral with respect to x over the given interval: 01(6x+15)dx =[3x2+15x]01=(3+15)(0) = 18
  3. Result:

    • The value of the triple integral V(x+y+z)dzdydxover the region V defined by 0x1, 0y2, and 0z3 is 18.