How do multiple integrals work?

Question

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?

Aurora Alvarado · Accepted Answer

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.

Aurora Alvarado · Answer

undefined Multiple integrals extend the concept of a single integral to integrate functions of more than one variable over regions in higher-dimensional spaces. They involve iterated integration, where you integrate one variable at a time, treating the others as constants. The basic idea is to slice the region of integration into small pieces, typically rectangles in two dimensions or parallelepipeds in three dimensions. You then sum up the contributions from each piece by taking limits of the Riemann sums as the size of the pieces approaches zero. In essence, multiple integrals allow you to find the volume, area, mass, center of mass, and other quantities for functions of multiple variables over specified regions in space.

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?

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?