Double IntegralsCalculus IV LabKaren E. Donnelly Saint Joseph's CollegeAll Rights Reservedrestart; with(plots): with(Student[MultivariateCalculus]);

We can compute double integrals as iterated integrals in Maple just as we would by hand: int(int(x^3*y^2,y=x^2..4),x=0..2);With the MultiInt command in the Student[MultivariateCalculus] package we can cut down on the typing and number of parentheses required:MultiInt(x^3*y^2,y=x^2..4,x=0..2);

Even Maple can struggle with double integrals!Sometimes Maple has the same problem we do when trying to perform the integration in a particular "given" order. For example, consider the iterated integral of the function NiMqJiUieEciIiQtJSRzaW5HNiMqJCUieUdGJSIiIw==over the region lying in the first quadrant bounded below by the parabola NiMvJSJ5RyokJSJ4RyIiIw==, and above by the horizontal line NiMvJSJ5RyIiJQ==. When we try to have Maple compute this integral, we have to wait a long time to find out that Maple can't do it.int(int(x^3*(sin(y^3))^2,y=x^2..4),x=0..2);However, reversing the order of integration works: NiMtSSRpbnRHNiRJKnByb3RlY3RlZEdGJkkoX3N5c2xpYkc2IjYkLUYkNiQqJkkieEdGKCIiJC1JJHNpbkdGJTYjKiRJInlHRihGLiIiIy9GLTsiIiEtSSVzcXJ0R0YlNiNGMy9GMztGNyIiJQ==. We can see why things get better by computing the inside integral first: insideint:=int(x^3*(sin(y^3))^2,x=0..sqrt(y));This has a computable antiderivative with respect to y:indefiniteintegral := int(insideint,y);Thus the final answer is now computatble by Maple: ANS := int(insideint,y=0..4);If we wish to have a 10 digit approximation to this exact value, thenevalf(ANS);The combined command would just be:int(int(x^3*(sin(y^3))^2,x=0..sqrt(y)),y=0..4) ; evalf(%);There are times when reversing the order of integration will not enable you or Maple to solve a particular double integral problem. When this happens, you must use numerical approximation method such as approximating Riemann sums.

The iterated integral in rectangular coordinates NiMtSSRJbnRHNiRJKnByb3RlY3RlZEdGJkkoX3N5c2xpYkc2IjYkLUYkNiQtSSJmR0YoNiRJInhHRihJInlHRigvRjA7L0YwLUkiZ0dGKDYjRi8vRjAtSSJoR0YoRjYvRi87L0YvSSJhR0YoL0YvSSJiR0Yo becomes the following iterated integral in polar coordianates: NiMtSSRJbnRHNiRJKnByb3RlY3RlZEdGJkkoX3N5c2xpYkc2IjYkLUYkNiQqJi1JImZHRig2JComSSJyR0YoIiIiLUkkY29zR0YlNiNJJnRoZXRhR0YoRjIqJkYxRjItSSRzaW5HRiVGNUYyRjJGMUYyL0YxOy9GMS1JInVHRihGNS9GMS1JInZHRihGNS9GNjsvRjYmRjY2I0YyL0Y2JkY2NiMiIiM= As an example of using Maple to solve integration problems using polar coordinates: NiMvJSJ4RyomJSJyRyIiIi0lJGNvc0c2IyUmdGhldGFHRic=, NiMvJSJ5RyomJSJyRyIiIi0lJHNpbkc2IyUmdGhldGFHRic=, consider the following problem: Integrate the function NiMvLSUiZkc2JCUieEclInlHLSUlc3FydEc2IywoIiIlIiIiKiRGJyIiIyEiIiokRihGMEYuover the region R which lies in the first quadrant bounded between the two circles: NiMvLCYqJCUieEciIiMiIiIqJCUieUdGJ0YoRig= and NiMvLCYqJCUieEciIiMiIiIqJCUieUdGJ0YoIiIl. and the lines NiMvJSJ5RyIiIQ== and NiMvJSJ5RyUieEc=. First we plot the function with slices to help visualize the portion of the surface we are considering for the volume under the surface over the region. f := (x,y) -> sqrt(4-x^2+y^2);Surf1 := plot3d(f(x,y),x=0..2, y = 0..2,axes=boxed):%;Slices := plot3d({1,2}, theta = 0..Pi/4,z=-1..2.5,coords=cylindrical,style=PATCHNOGRID,axes=boxed):%;display({Surf1, Slices},axes = BOXED);To visualize the region R of integration, we plot it using Maple and rectangular coordinates:P1 :=plot(sqrt(1-x^2),x=1/sqrt(2)..1):P2 :=plot(sqrt(4-x^2),x=sqrt(2)..2):P3 := plot(x,x=1/sqrt(2)..sqrt(2)): # The line y = xdisplay({P1,P2,P3},view = [0..2,0..2],scaling=constrained);Setting up this double iterated integral in Cartesian coodindates would be difficult because you would have to break the region R up into simpler subregions. It is very simple in polar coordinates. In polar coordinates the integral is the following: NiMtSSRpbnRHNiRJKnByb3RlY3RlZEdGJkkoX3N5c2xpYkc2IjYkLUYkNiQqJi1JJXNxcnRHRiU2IywmIiIlIiIiKiRJInJHRigiIiMhIiJGMkY0RjIvRjQ7RjJGNS9JJnRoZXRhR0YoOyIiISomSSNQaUdGJkYyRjFGNg==int(int(sqrt(4-r^2)*r,r=1..2),theta=0..Pi/4); simplify(%);or, in stages it would be: IndefiniteInner := int(sqrt(4-r^2)*r,r); Inner := int(sqrt(4-r^2)*r,r=1..2); Inner := simplify(Inner);IndefiniteOuter := int(Inner, theta); Outer:= int(Inner, theta = 0..Pi/4);The MultiInt command verifies our work. Note if we use the output = integral option -- the integral is shown unevaluated (this is called the inert form in Maple) -- to evaluate we would then use the value command. MultiInt(sqrt(4-r^2)*r,r=1..2,theta=0..Pi/4,output=integral);value(%);simplify(%);Leaving the output = integral options out we just get our answer which we simplify:MultiInt(sqrt(4-r^2)*r,r=1..2,theta=0..Pi/4); simplify(%);

1. Evaluate the double integral of NiMqJCUieUciIiM= over the region R in the first quadrant that is outside the circle NiMvJSJyRyIiIw== and inside the cardiod NiMvJSJyRyomIiIjIiIiLCZGJ0YnLSUkY29zRzYjJSZ0aGV0YUdGJ0Yn. Plot of the region R for this exercise: Portion in first quadrant outside the circle and inside the cartiod. polarplot({2, 2*(1+cos(theta))}, theta = 0..2*Pi);2. Find the volume V of the solid above the region R = {(NiQlInJHJSZ0aGV0YUc=)| NiMxIiIiJSJyRw== , NiMxJSJyRyIiJA==, NiQxIiIhJSZ0aGV0YUcxRiUqJiUjcGlHIiIiIiIlISIi } and under the surface NiMvJSJ6Ry0lJGV4cEc2IyomLCYqJCUieEciIiMiIiIqJCUieUdGLEYtRi0lI1BpRyEiIg==. Plot of the region R for this exercise -- portion of disk in first quadrant colored blue:P1 := polarplot({1, 3}, theta = 0..Pi/4, scaling=constrained, color = BLUE): P2 := polarplot({1, 3}, theta = Pi/4..2*Pi, scaling=constrained,color = RED): display({P1,P2});