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5.2 Diagram for the double integral

\curve(200,20,190,25) \curve(200,20,190,15) xx\curve(10,180,15,170) \curve(10,180,5,170) zz\curve(90,180,92,172) \curve(90,180,83,175) yy\curve(70,90,80,115,85,120,90,120,95,120,100,128,105,143,110,160) \curve(70,90,80,93,90,97,100,100,105,101.5,110,102.5,115,102,120,101,130,99,140, 96 ,150,93,160,91,170,90) \curve(170,90,175,93.5,176,94,178,95,180,96.5,185,102.5,186,104,187,106,190,112, 195,125,200,138,210,155) \curve(110,160,120,162,130,163,140,162,150,161,160,159,170,158,180,157,190, 155,200,154,210,155) \curve(150,140,160,137) \curve(150,140,155,150) \curve(155,150,165,147) \curve(160,137,165,147) To estimate the totalvolume under the surfacewe split the region RR up intosmaller rectangles RijR_{ij}, andapproximate the volume ofthe column above each one.If (xi,yj)(x_{i},y_{j}) is the south-westcorner of RijR_{ij}, then the volume ofthis column is roughly:f(xi,yj)hkf(x_{i},y_{j})hk.hhkk\curvedashes[1mm]1,0.5,1,0.5 \curve(155,139,155,125) \curve(150,115,155,125) \curve(165,125,155,125)