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Double, triple integration

  1. Jan 5, 2005 #1
    alo, i ve done single variable integration at school but i m trying to understand some vector analysis and going through books like Schey's Div, Grad Curl and all that as well as Schaum s vector analysis and to be able to understand i need to know exactly what a double integral is....i ve searched online and i can see you kind of integrate twice wrt the diff variables but if anybody could elaborate i d appreciate it, also can you have a double integral without limits?

    thanks in advance and sorry for the hassle

  2. jcsd
  3. Jan 5, 2005 #2


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    Recall that the definite integral of f (where f is a function of a single variable) from a to b is defined as:

    [tex]\int_a^bf(x)dx=\lim_{n \to \infty}\sum_{i=1}^nf(x_i^*)\Delta x[/tex]
    where [tex]\Delta x = (b-a)/n[/tex].
    (Usually this is arrived at by dividing the interval in subintervals etc)

    For a function of two variables, you can graph it as a surface. Now suppose f is a function of two variables defined on a closed rectangle
    [tex]R=[a,b] \times [c,d][/tex]
    You can apply the same sort of ritual as for the one variable case. Split the domain in small rectangles and choose a sample point [itex](x_{ij}^*,y_{ij}^*)[/itex] inside every rectangle.
    Then form the sum:
    [tex]\sum_{i=1}^m\sum_{j=1}^n f(x_{ij}^*,y_{ij}^*)\Delta A[/tex]
    where [itex]\Delta A = \Delta x \Delta y[/itex].
    If [itex]f(x,y)\geq 0[/itex] on and inside R, you can interpret this as an approx. to the volume under the surface.
    The definite integral of f over R is defined as:
    [tex]\int \int \limits_R f(x,y)dA=\lim_{m,n \to \infty} \sum_{i=1}^m\sum_{j=1}^nf(x_{ij}^*,y_{ij}^*)\Delta A[/tex]
    if this limit exists.

    In evaluating these integrals, use is made of a special case of Fubini's theorem:
    If [itex]f[/itex] is bounded on the rectangle [itex]R=[a,b] \times [c,d][/itex] and discontinuous only on a finite number of smooth curves, then:
    [tex]\int \int \limits_R f(x,y)dA=\int_a^b\int_c^df(x,y)dydx=\int_c^d\int_a^bf(x,y)dxdy[/tex]
    if the iterated integrals exist.

    You can see it as this. If you integrate wrt y first, then you have a function of x left. This can be viewed as the area of a cross-section of the desired volume in the plane through x, perendicular to the x-axis. Integrating this wrt y gives the volume under the curve.
    Last edited: Jan 5, 2005
  4. Jan 5, 2005 #3
    In my text, "Multivariable Calculus" by Larson Hostetler, and Edwards (possibly the best intro. to vector analysis yet), the definition of a double integral is

    *[my annote: more formally, [tex]f:\mathbb{R}^2\rightarrow\mathbb{R}[/tex]]

    That is the two-variable analogue of the single-variable Reimann sum. Basically, your can think of it as adding volumes over a grid of infinitisemal squares (parallel to the single-variable case, areas over infinitesemal 1d intervals).

    Here is a basic example of such an integral and how it is used:

    [tex]A={(x,y)|0\leq x \leq 1, 0\leq y \leq 3}[/tex] (rectangle with vertices (0,0) and (1,3))
    [tex]\int_{A}\int x^2 + 5y\ dx\ dy = \int_0^3\int_0^1 x^2 + 5y\ dx\ dy[/tex]
    [tex]=\int_0^3(\int_0^1 x^2 + 5y\ dx) dy[/tex]
    Key thing here is any in integrating with respect to x, any other variable (y) is effectively a constant:
    [tex]=\int_0^3(\frac{x^3}{3}+5xy)|_0^1 dy[/tex]
    [tex]=\int_0^3(\frac{1}{3}+5y)\ dy[/tex]

    And the indefinite integral would work similarly:

    [tex]\int\int x^2 + 5y\ dx\ dy = \int(\int x^2 + 5y\ dx) dy[/tex]
    [tex]= \int (\frac{x^3}{3} + 5xy)\ dy[/tex]
    [tex]= \frac{1}{3}x^3y + \frac{5}{2}xy^2[/tex]
    (this particular integral is valid over all of [tex]\mathbb{R}^2[/tex]

    Again, any variable not being integrated works like a constant.

    Note that the order of integration, whether you integrate with respect to y and then x, or x and then y, is 'arbitrary', and when both integrals are defined they give the same answer.

    The definite integral I showed you was easy, because the area of integration (parallel to 'interval of integration') was a rectangle. Often the area of integration doesn't have straight sides, it can be a triangle or a circular arc or a cycloid:

    [tex]\int_0^3 \int_0^{1-y} f(x) dx dy[/tex]

    This definite integral is over a triangle; the inner integrand integrates over x from '0' to '1-y'. A picture would help here...

    In tricky cases like circles you can substitue variables, for instance [tex]\theta[/tex] and r for x and y. (this I found was quite hard)

    Triple integration is parallel, and you can actually define it inductively from the double integral.

    Let me plug my 3rd semester calc. book here, 'multivariable calculus' by Larson, Hostetler and Edwards - it's really clear and has tons of awesome graphics generated by Mathematica.
    Last edited by a moderator: Jan 5, 2005
  5. Jan 5, 2005 #4
    Well, Galileo beat me to it! :yuck:
  6. Jan 6, 2005 #5
    thanks a lot guys, much appreciated...i think i got enough of the general picture to have an idea of what is happening....i ll try playing around with some easy ones.....i ll try get my hands on the pictures from the "multivariable calculus" even just to get some mental images hehe

    thanks again

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