18 02 Multivariable Calculus Spring 2009 Lecture 17 Double and iterated integrals in the plane March 13 Reading Material From Simmons 20 1 and 20 2 From Course Notes I Last time Non independent variables Partial differential equations Today Double and iterated integrals in the plane 2 Integrals of functions of one variable Consider a function y f x such that x is in a b Assume also in this case that f x 0 on a b Then we know that the area A of the sub graph is given by Z Area A b f x dx a We also know how to evaluate this integral using an anti derivative F x F 0 x f x In fact by the fundamental theorem of calculus Z b f x dx F b F a A a A couple of questions are natural at this point Question Why areas are computed through integrals What does an integral really mean Take the interval a b and subdivide it into smaller intervals by introducing a partition a x0 x1 xi xn b 1 Call Ii xi xi 1 and xi xi 1 xi For each interval Ii we pick a point x i in it and we measure the hight hi f x i Then if Ri is the rectangle of base Ii and hight hi it is clear that the area of the sub graph A is approximated by the sum of the areas of the rectangles Ri s n 1 X A An f x i xi i 1 If we consider partitions that contain more and more points e i finer ones then we have A lim An lim n n n 1 X Z f x i xi b f x dx a i 1 and the last term is a purely symbolic definition 3 Iterated Integrals Consider now a function of two variables z f x y defined on a rectangle a b c d 2 Then the volume V of the sub graph of f x y above the rectangle R is Z Z V Z f x y dA R a b Z d f x y dy dx c z inner integral z outer integral this is a double integral defined as iterated integrals We will go back to the method of evaluation later now we discuss why one can compute volumes with double integrals As before we consider a partition for a b and we also consider a partition for c d c y0 y1 yj ym d and we define the intervals Jj yj yj 1 We then define the rectangles Ri j Ii Ij of area Aij xi yj we pick a point P i j in Ri j and we define the hight hi j f P i j Now it is easy to see that the volume V is approximated by the sum of the volumes of the parallelepipeds defined by Ri j hi j and hence V Vn m n 1 X m 1 X i 1 j 1 3 f P i j xi yj Now as we make the partitions finer and finer we have V lim lim Vn m lim lim n m 4 n m n 1 X m 1 X Z Z f P i j xi yj f x y dA R i 1 j 1 Method of evaluation Given a function z f x y for x y in R a b c d we compute the double interval Z Z V f x y dA R by using the method of iterated integrals 1 We fix one of the variable say x in a b that we call outer variable and perform the inner integral Z d f x y dy A x c This function A x represents the area of the intersection of the solid given by the sub graph of f with the plane parallel to the yz plane and through the point x 0 0 2 Now to get the volume V of the solid given by the sub graph we need to integrate all these areas A x along a b hence we perform the outer integral Z Z Z b V b d A x dx a f x y dy dx a 4 1 c Remark When the domain of integration is a rectangle the choice of the outer or inner variable is totally arbitrary In fact above we could have picked y as the outer variable and replaced 4 1 by Z Z d b V f x y dx dy c a In this case we say that we changed the order of integration 4 4 2 Exercise 1 Consider the function f x y 20 x2 y 2 Compute the volume of the solid under the graph of f and above the rectangle R 1 2 1 3 Solution 5 Double integrals with non rectangular basis We start with the following problem Exercise 2 Consider again the function f x y 20 x2 y 2 as above Compute the volume of the solid below the graph of f and above the triangle of vertices 1 1 2 1 and 2 3 5 Solution The procedure is similar to the one described above with the iterated integrals The problem is that it is more tricky to find the limits of integration 1 Fix the outer variable with respect to this variable the outer integral is done on the full range 2 Set up the inner integral where now the range depends on the outer variable see picture below In our resolution we fix x as the outer variable and its range is 1 2 then Inner Integral Outer Integral 6 Study Guide 1 Answer the following questions I explained what is the geometric picture behind formula 4 1 Coud you explain in a similar way the geometric meaning of 4 2 If you compare 4 1 and 4 2 you see that changing the order of integration doesn t change anything for the limits of integration for the respective variables Is this also true for Exercise 2 7
View Full Document
Unlocking...