18 02A Topic 31 Double and iterated integrals Author Jeremy Orloff Read TB 20 1 20 2 SN I 1 Double and iterated integrals are really the same thing The distinction is that a double integral is geometric i e a slicing and summing and an iterated integral is analytic i e a computational tool Iterated integrals Z 1Z 2 xy dx dy Example 0 0 Z 2 2 x2 xy dx Inner integral y 2y 2 0 0 Z 1 1 Outer integral 2y dy y 2 0 1 0 Z 1 Z x 2x 2y dy dx Example 0Z x x2 2x 2y dy 2xy y 2 Inner integral x2 Z Outer integral x x2 1 3x2 2x3 x4 dx x3 0 2x2 x2 2x3 x4 3x2 2x3 x4 x4 x5 2 5 1 1 0 Double integrals Mass example 1 dimension We start with a single one dimensional integral for mass density x units are mass length dm x dx Z Z b M dm x dx 1 1 2 5 dx a b a y 2 dimensions Now we consider a double two dimensional integral for mass density x y units are mass area dm x y dA x y dx dy Z Z Z Z M x y dA x y dx dy R R This is a sum over the infinitesimal boxes with area dx dy x area dx dy continued 1 18 02A topic 31 2 To compute this sum we ll learn to do an iterated integral Let s discover this for ourselves in the simple example of the rectangle shown The box runs between x 1 and x 3 and y 1 and y 4 We ll assume a variable density x y x2 y To find the mass we have to sum over all the little squares The mass in a single square at coordinates x y is y 4 dm x y dx dy To do the sum we first fix y and then sum the mass of the boxes over the horizontal strip see the figure This gives the mass of the strip is Z 3 Z 3 x2 y dx dy x y dx dy 1 1 x3 y dy 3 3 1 26 y dy 3 y 1 x 1 Notice that the integral i e the sum along the horizontal strip depends on y different strips will have different mass Next to find the total mass we Z 4have to sum up the mass of all the horizontal strips 26 15 195 26 y dy This is just an integral in y 3 2 3 1 3 We followed one integral by another that is we computed the iterated integral Z 4Z 3 Z 4Z 3 x y dx dy x2 y dx dy 1 1 1 3 1 Limits of integration To compute the iterated integral over more complicated regions we need to be able to find limits of integration We start with an example y x 1 Example Find the mass of region R bounded by y x 1 y x2 x 0 x 1 density x y xy Inner limits y from x2 to x 1 Outer limits Z Z x from 0 to 1 Z Z 1 M x y dA R Inner xy dy x x2 1 Z Outer 0 y 2 x2 y x2 1 x x 1 2 x5 x3 x x5 x2 2 2 2 2 2 x3 x x5 x4 x3 x2 x6 x2 dx 2 2 2 8 3 4 12 2 y x2 xy dy dx x 0 2 x 1 x 1 Z R x 1 1 0 1 1 1 1 5 8 3 4 12 8 Note The syntax y x in limits is redundant but useful We know it must be y because of the dy matching the integral sign continued 18 02A topic 31 3 Volume Like area volume sum of rectangular boxes y 1 z f x y 1 1 1 Tetrahedron dV z dA y 1 x R 1 x Region R Example Volume of tetrahedron see above pictures Surface z 1 x y Z 1 Z 1 x Limits inner 0 y 1 x outer 0 x 1 V 1 x y dy dx x 0 y 0 1 x 1 x 1 x2 y2 1 x x x2 x 1 x y dy y xy Inner 2 0 2 2 y 0 Z 1 2 x 1 1 1 1 1 x dx Outer 2 2 2 6 6 0 2 Z Note Finding volumes is nice But just like single integrals and area it is only one way to think of integrals The most important way is still to think of them as a sum Changing order of integration Z 1Z 1 y e Example dy dx Inner integral is too hard so change order y 0 x 1 Find limits for region R inner 0 y x outer 0 x 1 inner 0 x y 2 outer 0 y 1 Z 1 Z y2 y e 3 Compute integral dx dy y 0 x 0 y 2 Reverse limits ey Inner x y y2 y ye 0 Z Outer 0 1 yey yey ey 10 1 y 1 y x or x y 2
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