Question

How do multiple integrals work?

When I try to mentally calculate a double, triple, or quadruple integral, my answers are always different from what my calculator presents.
I can calculate definite integrals easily `int_b^a` `f(c)dx` `=` `(a)(c)` `-` `(b)(c)`
but how do other integrals work and how are they calculated?

Answer 1

It's a bit like the way partial derivatives work where you treat other variables as constant and perform the derivative against a particular variable.

So for Partial derivatives:
Eg. `partial/(partialx) xy^2 = y^2` because we treat `y` as constant and so we perform `d/dx (ax) = a`

For a traditional single definite integral we are summing up infinitesimal vertical bars to find an area.

For a double integral we have something like

`int int _R f(x,y) dA`

where `R` is called the region of integration and is a region in the `(x, y)` plane. The double integral gives us the volume under the surface `z = f(x, y)`, just as a single integral gives the area under a curve.

To evaluate a double integral we do it in stages, starting from the inside and working out, using our knowledge of the methods for single integrals. The easiest kind of region `R` to work with is a rectangle, but it can be any region or line (for a line integral) or a closed loop where we put a circle on the integral as in `oint_R f(x,y)dS`

E.g. If `f(x,y)=1+8xy` and we want to integrate over the region bounded by `0 le x le 3` and `1 le y le 2` then we would have:

`int int_R f(x,y) dA = int_1^2 int_0^3 (1+8xy) dx dy`

Or to be more explicitly;

`int int_R f(x,y) dA = int_(y=1)^(y=2) int_(x=0)^(x=3) (1+8xy) dx dy`

We evaluate the "inner integral" by treating `y` as constant so:

`int int_R f(x,y) dA = int_(y=1)^(y=2) {int_(x=0)^(x=3) underbrace((1+8xy) dx)_("treat y as constant")} dy`

`" " = int_(y=1)^(y=2) {[x+(8x^2y)/2]_(x=0)^(x=3)} dy`

`" " = int_(y=1)^(y=2) {[x+4x^2y]_(x=0)^(x=3)} dy`

`" " = int_(y=1)^(y=2) {(3+4*9*y)-(0+0)} dy`

`" " = int_(y=1)^(y=2) {3+36y} dy`

`" " = [3y+(36y^2)/2]_(y=1)^(y=2)`

`" " = [3y+18y^2]_(y=1)^(y=2)`

`" " = (3*2+18*4) - (3*1+18*1)`
`" " = (6+72) - (3+18)`
`" " = 78 - 21`
`" " = 57`

Hope that helps. Feel free to ask for further help or examples.