18.02A Topic 31: Double and iterated integrals.Read: TB: 20.1, 20.2Double integral –geometricIterated integral –analyticIterated integralsExample:Z10Z20xy dx dyInner integral:Z20xy dx =x22y20= 2yOuter integral:Z102y dy = y210= 1.Example:Z10Zxx2(2x + 2y) dy dxInner integral:Zxx2(2x + 2y) dy = 2xy + y2xx2= 2x2+ x2− (2x3+ x4) = 3x2− 2x3− x4Outer integral:Z103x2− 2x3− x4dx = x3−x42−x5510= 1 −12−15.Double integrals Mass example1 dimension:density = δ(x)dm = δ(x) dxM =Zbaδ(x) dxabdx2 dimensions:density = δ(x, y)dm = δ(x, y) dA = δ(x, y) dx dyM =Z ZRδ(x, y) dA =Z ZRδ(x, y) dx dy= Double integralTo compute: mass (horiz.) strip (fix y) =Zbaδ(x, y) dxM = ”sum” mass of strips =ZdcZbaδ(x, y) dx dy= iterated integral.abcddxdy(continued)118.02A topic 31 2Limits of integrationExample: Find the mass of region R bounded byy = x + 1, y = x2, x = 0, x = 1, density = δ(x, y) = xyx is between 0 and 1.As x moves the vertical lines sweep out R.Fix x then y runs from x2to x + 1.⇒ M =Z ZRδ(x, y) dA =Z1x=0Zx+1y=x2xy dy dxy = x + 1y = x21RInner:Zx+1x2xy dy = xy22x+1x2=x(x + 1)22−x52=x32+ x2+x2−x52Outer:Z10x32+ x2+x2−x52dx =x48+x33+x24−x61210=18+13+14−112=58.Note: The syntax y = x2in limits is redundant but useful. We know it must be ybecause of the dy matching the integral sign...Volume: Like area: volume = ’sum’ of rectangular boxes.tttttt\\\\\\\tttttt\\\\\\\tt\\tt\\dV = z dAkkkkkkkkkkkk??????????111z = f(x, y) Tetrahedron??????????11xyRy = 1 − xRegion RExample: Volume of tetrahedron (see above pictures)Surface: z = 1 − x − y.Limits: x: 0 to 1; y: 0 to 1 − x.⇒ V =Z1x=0Z1−xy=01 − x − y dy dx.Inner:Z1−xy=01 − x − y dy = y − xy −y221−x0= 1 − x − x + x2−12+ x −x22Outer:Z1012− x +x22dx =12−12+16=16.Changing order of integrationExample:Z10Z1√xeyydy dx. –Inner integral is too hard –so change order:1) Find limits for region R: x from 0 to 1; fix x: y from√x to 1.2) Reverse limits: y from 0 to 1; fix y: x from 0 to y2.3) Compute integral:Z1y=0Zy2x=0eyydx dyy = 1y =√x or x = y2Inner: xeyyy20= yey⇒ Outer:Z10yey= yey− ey|10=
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