MIT OpenCourseWare http ocw mit edu 18 02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use visit http ocw mit edu terms I Limits in Iterated Integrals For most students the trickiest part of evaluating multiple integrals by iteration is to put in the limits of integration Fortunately a fairly uniform procedure is available which works in any coordinate system You must always begin by sketching the region in what follows we ll assume you ve done this 1 Double integrals in rectangular coordinates Let s illustrate this procedure on the first case that s usually taken up double integrals in rectangular coordinates Suppose we want to evaluate over the region R pictured the integral R region between x2 y2 1 and x y 1 we are integrating first with respect to y Then to put in the limits 1 Hold x fixed and let y increase since we are integrating with respect to y As the point x y moves it traces out a vertical line 2 Integrate from the y value where this vertical line enters the region R to the y value where it leaves R 3 Then let x increase integrating from the lowest x value for which the vertical line intersects R to the highest such x value Carrying out this program for the region R pictured the vertical line enters R where y 1 x and leaves where y d m I A The vertical lines which intersect R are those between x 0 and x 1 Thus we get for the limits JJ f x Y dydx To calculate the double integral integrating in the reverse order J f x y dx dy 1 Hold y fixed let x increase since we are integrating first with respect to x This traces out a horizontal line 2 Integrate from the x value where the horizontal line enters R to the x value where it leaves 3 Choose the y limits to include all of the horizontal lines which intersect R Following this prescription with our integral we get x l y JJ f x Y dx dy Exercises 3A 2 1 18 02 NOTES 2 2 Double integrals in polar coordinates The same procedure for putting in the limits works for these integrals also Suppose we want to evaluate over the same region R as before As usual we integrate first with respect to r Therefore we 1 Hold B fixed and let r increase since we are integrating with respect to r As the point moves it traces out a ray going out from the origin 2 Integrate from the r value where the ray enters R to the r value where it leaves This gives the limits on r 3 Integrate from the lowest value of 8 for which the corresponding ray intersects R to the highest value of 8 To follow this procedure we need the equation of the line in polar coordinates We have x y l rcose rsinB l or r 1 cose This is the r value where the ray enters the region it leaves where r 1 The rays which intersect R lie between B 0 and B 2 Thus the double iterated integral in polar coordinates has the limits Lm 2 p sine r l cose sine dr dB L m s e sin e Exercises 3B 1 3 Triple integrals in rectangular and cylindrical coordinates You do these the same way basically To supply limits for J dz dy dx over the region D we integrate first with respect to z Therefore we 1 Hold x and y fixed and let z increase This gives us a vertical line 2 Integrate from the z value where the vertical line enters the region D to the z value where it leaves D 3 Supply the remaining limits in either xy coordinates or polar coordinates so that you include all vertical lines which intersect D This means that you will be integrating the remaining double integral over the region R in the xy plane which D projects onto For example if D is the region lying between the two paraboloids 2 z x y we get by following steps 1 and 2 2 z 4 x 2 y2 I LIMITS IN ITERATED INTEGRALS 3 where R is the projection of D onto the xy plane To finish the job we have to determine what this projection is From the picture what we should determine is the xy curve over which the two surfaces intersect We find this curve by eliminating z from the two equations getting x2 y2 4 x2 Y 2 x2 Y2 which implies 2 Thus the xy curve bounding R is the circle in the xy plane with center a t the origin and radius fi This makes it natural to finish the integral in polar coordinates We get the limits on z will be replaced by r2 and 4 r2 when the integration is carried out Exercises 5A 2 4 Spherical coordinates Once again we use the same procedure To calculate the limits for an iterated integral J dpd4d6 over a region D in bspace we are integrating first with respect to p Therefore we 1 Hold the origin 4 and 6 fixed and let p increase This gives us a ray going out from 2 Integrate from the p value where the ray enters D to the p value where the ray leaves D This gives the limits on p 3 Hold 6 fixed and let 4 increase This gives a family of rays that form a sort of fan Integrate over those values for which the rays intersect the region D 4 Finally supply limits on 6 so as to include all of the fans which intersect the region D For example suppose we start with the circle in the yz plane of radius 1 and center at l O rotate it about the z axis and take D to be that part of the resulting solid lying in the first octant First of all we have to determine the equation of the surface formed by the rotated circle In the yz plane the two coordinates p and 4 are indicated To see the relation between them when P is on the circle we see that also angle OAP 4 since both the angle 4 and OAP are complements of the same angle AOP From the right triangle this shows the relation is p 2 sin 4 As the circle is rotated around the z axis the relationship stays the same so p 2 sin 4 is the equation of the whole surface To determine the limits of integration when 4 and 6 are fixed the correpsonding ray enters the region where p 0 and leaves where p 2 sin 4 As 4 increases with 6 fixed it is the rays between 4 0 and 4 r 2 that intersect D since we are only considering the portion of the surface lying in the first octant and thus above the xy plane 4 18 02 NOTES Again since we only want the part in …
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