Dr. Z’s Math251 Handout #15.2 [Iterated Integrals]By Doron ZeilbergerProblem Type 15.2a: Calculate the iterated integralZbaZdcf(x, y) dx dy .Example Problem 15.2a: Calculate the iterated integralZ41Z20(x +√y) dx dy .Steps Example1. First evaluate the inner integral.Zdcf(x, y) dx .Since it is an x-integral, integrate with re-spect to x, and treat y as a constant. Theanswer to this step should be an expres-sion in y. Of course, once in awhile itcould be just a number.1. The inner integral isZ20(x +√y) dxThe anti-derivative is x2/2 + x√y, soZ20(x +√y) dx =x22+ x√y20=22− 022+ (2 − 0)√y = 2 + 2y1/2.2. Do the outer integral by integratingw.r.t. y the expression that you got instep 1.2.Z41Z20(x +√y) dx dy=Z41Z20(x +√y) dxdy=Z41[2 + 2y1/2] dy = 2y + 2y3/23/241= 2y+43(√y)341= 2·(4−1)+43[(√4)3−(√1)3]= 6 +43· 7 =463.Ans.: 46/3.1Problem Type 15.2b: Calculate the double integralZ ZRf(x, y) dA ,R = {(x, y) |a ≤ x ≤ b , c ≤ y ≤ d } .Example Problem 15.2b: Calculate the double integralZ ZRxy2x2+ 1dA ,R = {(x, y) |0 ≤ x ≤ 1 , −3 ≤ y ≤ 3 } .Steps Example1. Convert it to an iterated integral:ZbaZdcf(x, y) dy dx ,or, if you wishZdcZbaf(x, y) dx dy .Both are correct.1.Z10Z3−3xy2x2+ 1dy dx .22. Evaluate the iterated intergal like in15.2a .2.Z10Z3−3xy2x2+ 1dy dx=Z10Z3−3xy2x2+ 1dydx .The inner integral is:Z3−3xy2x2+ 1dy =xx2+ 1Z3−3y2dy =xx2+ 1y333−3=xx2+ 118 =18xx2+ 1.Hence the iterated integral is:Z1018x(x2+ 1)dx = 9 ln(x2+ 1)10= 9[ln(12+ 1) − ln(02+ 1)] = 9 ln 2 .Ans.: 9 ln 2.Problem Type 15.2c: Find the volume of the solid that lies under the plane ax + by + cz = dand above the rectangleR = {(x, y) |A1≤ x ≤ A2, B1≤ y ≤ B2} .Example Problem 15.2c: Find the volume of the solid that lies under the plane 2x +3y +z = 10and above the rectangleR = {(x, y) |0 ≤ x ≤ 2 , 0 ≤ y ≤ 1} .Steps Example1. Solve for z putting it in the form z =f(x, y). Technically, you would have tocheck that f(x, y) is always positive abovethe given region R but you can trust theproblem. Set up the integralZRf(x, y) dA .1. Solving for z in 2x + 3y + z = 10 givesz = 10−2x−3y, so f(x, y) = 10−2x−3yand the desired volume isZR(10 − 2x − 3y) dA .32. Convert the area-integral to an iter-ated integral either way.2.ZR(10−2x−3y) dA =Z10Z20(10−2x−3y) dx dy .Or, if you wishZR(10−2x−3y) dA =Z20Z10(10−2x−3y) dy dx .3. Compute (one of those) iterated inte-gral(s), like we did in 15.2a.3.Z10Z20(10−2x−3y) dx dy =Z10Z20(10 − 2x − 3y) dxdy=Z1010x − x2− 3xy20dy =Z1010 · 2 − 22− 3 · 2y − 0 dy=Z10[16 − 6y] dy= 16y − 3y210= 16 − 3 = 13 .Ans.: The volume is 13.Problem Type 15.2d: Find the volume of the solid in the first octant bounded by the cylinderz = r2− x2and the plane y = a.Example Problem 15.2d: Find the volume of the solid in the first octant bounded by the cylinderz = 16 −x2and the plane y = 3.Steps Example41. Unlike the previous problem, now wealso need to figure out the “floor plan”R. If the surface is z = f(x, y) then setf(x, y) = 0 getting x = ±r and rememberthat positive octant means x ≥ 0, y ≥ 0.It follows thatR = {(x, y)|0 ≤ x ≤ r, 0 ≤ y ≤ a} .1. Setting z = 0 gives 16 − x2= 0 sox = ±4 but in the positive orthant wehave x ≥ 0 so the boundary of R are x =0, x = 4, y = 0, y = 3. So the boundaryis:R = {(x, y)|0 ≤ x ≤ 4, 0 ≤ y ≤ 3} .2. Set up the integralZRf(x, y) dA ,and convert it to an iterated integral.2.ZR(16−x2) dA =Z30Z40(16−x2) dx dy .3. Evaluate this iterated integral. 3.Z30Z40(16−x2) dx dy =Z30Z40(16 − x2) dxdy=Z3016x −x3340dy =Z301283dy=1283y30= 128 .Ans.: The volume is