How do multiple integrals work?

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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 $\int_{b}^{a}$ $int_b^a$ $f (c) d x$ $f(c)dx$ $=$ $=$ $(a) (c)$ $(a)(c)$ $-$ $-$ $(b) (c)$ $(b)(c)$
but how do other integrals work and how are they calculated?

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Answer 1 · 2016-12-18T20:39:48

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. $\frac{\partial}{\partial x} x y^{2} = y^{2}$ $partial/(partialx) xy^2 = y^2$ because we treat $y$ $y$ as constant and so we perform $\frac{d}{d x} (a x) = a$ $d/dx (ax) = a$

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

$\int \int_{R} f (x, y) d A$ $int int _R f(x,y) dA$

where $R$ $R$ is called the region of integration and is a region in the $(x, y)$ $(x, y)$ plane. The double integral gives us the volume under the surface $z = f (x, y)$ $z = f(x, y)$ , 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 $R$ $R$ 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 $\oint_{R} f (x, y) d S$ $oint_R f(x,y)dS$

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

$\int \int_{R} f (x, y) d A = \int_{1}^{2} \int_{0}^{3} (1 + 8 x y) d x d y$ $int int_R f(x,y) dA = int_1^2 int_0^3 (1+8xy) dx dy$

Or to be more explicitly;

$\int \int_{R} f (x, y) d A = \int_{y = 1}^{y = 2} \int_{x = 0}^{x = 3} (1 + 8 x y) d x d y$ $int int_R f(x,y) dA = int_(y=1)^(y=2) int_(x=0)^(x=3) (1+8xy) dx dy$

We evaluate the "inner integral" by treating $y$ $y$ 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.

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How do multiple integrals work?

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