18.02A Topic 31: Double and iterated integrals.Author: Jeremy OrloffRead: TB: 20.1, 20.2, SN: I.1Double and iterated integrals are really the same thing. The distinction is that adouble integral is geometric, i.e. a slicing and summing, and an iterated integral isanalytic, i.e. a computational tool.Iterated 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:We start with a single (one dimensional) integral for mass:density = δ(x) units are mass/lengthdm = δ(x) dxM =Zdm =Zbaδ(x) dxabdx2 dimensions:Now we consider a double (two dimensional) integral for massdensity = δ(x, y) units are mass/areadm = δ(x, y) dA = δ(x, y) dx dy⇒ M =Z ZRδ(x, y) dA =Z ZRδ(x, y) dx dy.This is a ’sum’ over the infinitesimal boxes with area dx dy.xyarea = dx dy(continued)118.02A topic 31 2To compute this ’sum’ we’ll learn to do an iterated integral. Let’s discover this forourselves in the simple example of the rectangle shown. The box runs between x = 1and x = 3 and y = 1 and y = 4. We’ll assume a variable density δ(x, y) = x2y.To find the mass we have to ’sum’ over all the little squares.The mass in a single square at coordinates (x, y) isdm = δ(x, y) dx dy.To do the ’sum’ we first fix y and then ’sum’ the mass of theboxes over the horizontal strip (see the figure). This gives themass of the strip isZ31δ(x, y) dx dy =Z31x2y dx dy =x33y dy31=263y dy.Notice that the integral (i.e. the sum along the horizontalstrip) depends on y –different strips will have different mass.1 314xyyNext, to find the total mass we have to sum up the mass of all the horizontal strips.This is just an integral in y:Z41263y dy =263·152=1953.We followed one integral by another –that is we computed the iterated integralZ41Z31δ(x, y) dx dy =Z41Z31x2y dx dy.Limits of integrationTo compute the iterated integral over more complicated regions we need to be ableto find limits of integration. We start with an example.Example: Find the mass of region R bounded byy = x + 1, y = x2, x = 0, x = 1, density = δ(x, y) = xy.Inner limits: y from x2to x + 1.Outer limits: x from 0 to 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...(continued)18.02A topic 31 3Volume: Like area: volume = ’sum’ of rectangular boxes.dV = z dA111z = f (x, y)Tetrahedron11xyRy = 1 − xRegion RExample: Volume of tetrahedron (see above pictures)Surface: z = 1 − x − y.Limits: inner: 0 < y < 1 − x, outer: 0 < x < 1. ⇒ 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.Note: Finding volumes is nice. But just like single integrals and area, it is only oneway to think of integrals. The most important way is still to think of them as a ’sum’.Changing order of integrationExample:Z10Z1√xeyydy dx. –Inner integral is too hard –so change order:1) Find limits for region R: inner: 0 < y <√x, outer: 0 < x < 1.2) Reverse limits: inner: 0 < x < y2, outer: 0 < y < 1.3) Compute integral:Z1y=0Zy2x=0eyydx dyy = 1y =√x or x = y2Inner: xeyyy20= yey⇒ Outer:Z10yey= yey− ey|10=
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