cyjdaxibxy(x*i j,y*i j)Figure13.1.The sample points and solid volume of a Riemann sum3.Iff(x,y)≥0 onR, thenRRRf(x,y)dArepresents the volume of the solid whose upperboundary is the surfacez=f(x,y) and whose lower boundary isR. That is,ZZRf(x,y)dA=vol. (x,y,z) : (x,y)∈R,0≤z≤f(x,y)=vol. (x,y,z) :a≤x≤b,c≤y≤d,0≤z≤f(x,y).4. From now on, we shall use the notation|D|to denote the area of a plane regionD. Let usdivide both sides of (13.1) by|R|=(b-a)(c-d). SinceΔA=|R|/n2, we haveΔA/|R|=1/n2, sowe get1|R|ZZRf(x,y)dA=limn→∞1n2nXi=1nXj=1f(x*i j,y*i j).Since1n2∑i∑jf(x*i j,y*i j) is the average of the sample valuesf(x*i j,y*i j), we interpret the last equa-tion in terms of the average value off:average(x,y)∈Rf(x,y)=1|R|ZZRf(x,y)dA.(13.2)Example13.1. Evaluate the double integral of the functionf(x,y)=(5-2x-3yif 5-2x-3y≥0,0if 5-2x-3y≤0,over the rectangleR=[0,3]×[0,2].Solution. We shall use the interpretation of the double integral as volume. By Remark 3 above,the value of the integralZZRf(x,y)dAis the volume of the solidEthat lies betweenRand the part of the surfacez=f(x,y) with (x,y)∈R.The graph off(x,y) (before the restriction toR) consists of the part of the planez=5-2x-3ywith 5-2x-3y≥0 and the part of thexy-planez=0 with 5-2x-3y≤0. The line‘with equation5-2x-3y=0 splits the rectangleRin two parts: a triangle with vertices (0,0), (52,0), and (0,53),102
(0,0)(3,0)(3,2)(0,2)(52,0)(0,53)‘Figure13.2.The pyramidEand its basewhere 5-2x-3y≥0; and a pentagon with vertices (52,0), (3,0), (3,2), (0,2), and (0,53), where5-2x-3y≤0 (see Figure 13.2). For points (x,y) lying in the pentagon, the graphz=f(x,y) andthexy-plane coincide, and for (x,y) in the triangle, the graphz=f(x,y) is a triangle in the planez=5-2x-3y. Thus, the solidEis the pyramid with verticesO(0,0,0),A(52,0,0),B(0,53,0), andC(0,0,5). Its volume is13|OC| ·area(4OAB)=13|OC| ·12|OA| |OB|=135·125253=12536.Hence,ZZRf(x,y)dA=12536.13.2. Iterated integralsIn practice, we want to reverse the order of things in the above example, so that we are able tocompute volumes, averages, etc. by interpreting them as double integrals and then evaluating thelatter using some “standard” machinery independent of our interpretation. One such method is themethod of iterated integrals. Aniterated integralis an expression of one of the formsZdcZbaf(x,y)dx dyorZbaZdcf(x,y)dy dx.Sometimes, we write these integrals asZdcZbaf(x,y)dxdyandZbaZdcf(x,y)dydx,respectively. Note that these are definite integrals of functions defined by means of definite inte-grals andnotdouble integrals. Thus, we can evaluate such integrals using the familiar methodsfrom single-variable calculus.