Assume thatf(x, y) =zis the graph of a continuous (except possibly on points in thefinite union of graphs) non-negative function. ThenR RRfdxdyis the volume of the solidabove thez= 0 plane and below the graph off.Here is a fancier case. How do we find the volume under the functionf(x, y), abovethe planez= 0 and within the triangle with vertices (0,0),(1,0),(1,1).Apparently this isnot covered by the above theorems. We extend the functionfby making it equal to zerooutside the triangle and using the rectangle [0,1]×[0,1].The singularities are all on theliney=xand hence are allowable.How do we compute these double integrals.These integrals are the limits of sumsin a rectangular array.Here are two ways of organizing the summing of numbers in arectangular array.The first method is to add the numbers in each column and thenadding these sums. A second method is to add the numbers in each row and then addingthese sums.Formally we havei,j=nXi,j=1f(xi, yj)ΔxiΔyj=j=nXj=1i=nXi=1f(xi, yj)Δxi!Δyj=i=nXi=1j=nXj=1f(xi, yj)Δyj!Δxi.If we take the limit asn→ ∞, then these three sums have the same limit providedfis bounded, and continuous outside a finite union of graphs. This is a delicate result.Theorem3.Assume thatfis nice on a rectangleR= [a, b]×[c, d].The first sumconverges toZ ZRfdxdy.The second converges toZdc(Zbafdx)dy.The third converges toZba(Zdcfdx)dy.These are all equal.Example4.LetR= [-1,2]×[3,4], f(x, y) = 2x2+xy+y2.ThenZ ZR=Z43(Z2-1f(x, y)dx)dy.We separate out the inside integral to getZ2-1(2x2+xy+y2)dx=x3+x2y+xy2|x=2x=-1= 3y2+ (3/2)y+ 9.