CAS Exercise Examples for Chapter 15: Multiple Integrals

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1 CAS Exercise Examples for Chapter 5: Multiple Integrals à Section 5. Double Integrals 3 x Example: Integrate Ÿ Ÿ ÅÅÅÅÅÅ d dx using Mathematica. x The easiest wa to compute double integrals is to open the BasicInput palette and then click on the button containing Ÿ Ñ Ñ Ñ Ñ. Enter this twice and insert the appropriate values or functions in the positions indicated. You can also use the Integrate command as followsmathematica compute the integral. Note the difference in the order in which ou position the bounds for each example. In[8]:= Out[83]= Out[8]= Clear@x,, fd f@x_, _D := êhx L 3 x f@x, D x Integrate@f@x, D, 8x,, 3<, 8,, x<d Log@3D Log@3D To reverse the order of integration, it is best to first plot the region over which the integration extends. This can be done with ImplicitPlot and all bounds involving both x and can be plotted. A graphics package must be loaded. Remember to use the double equal sign for the equations of the bounding curves. In[85]:= << Graphics`ImplicitPlot` Now we will plot the region over which the integral extends.

2 CAS Exercise Examples for Chapter 5 In[86]:= ImplicitPlot@8, x, x, x 3<, 8x,, 3<, 8,, 3<, AxesLabel 8x, <D; x Another wa to plot this region is with the FilledPlot command. This requires a package to be loaded first. In[87]:= << Graphics`FilledPlot` This will shade the region between = and = x. In[88]:= FilledPlot@8x, <, 8x,, 3<D; To reverse the order of integration, we see that x will go from to 3, while goes from to 3.

3 CAS Exercise Examples for Chapter In[89]:= f@x, D x Out[89]= Log@3D Sometimes the reversal of the order of integration requires two integrals. Here, we will use both the Integrate command and the numerical NIntegrate command and also draw the region over which the integral extends. In[9]:= Out[9]= Clear@x,, fd f@x_, _D := x Integrate@f@x, D, 8,, <, 8x,, <D NIntegrate@f@x, D, 8,, <, 8x,, <D ImplicitPlot@8x, x,, <, 8x,, <, 8,,.<, AxesLabel 8x, <D; I + è!!! π HErfi@D Erfi@DLM Out[93]= x This integral does not have a closed form; the result is the Gaussian error function (from the normal distribution To reverse the order of integration for this function, two separate integrals must be used. In[95]:= Integrate@f@x, D, 8x,, <, 8,, x ê <D + Integrate@f@x, D, 8x,, <, 8,, <D NIntegrate@f@x, D, 8x,, <, 8,, x ê <D + NIntegrate@f@x, D, 8x,, <, 8,, <D Out[95]= H + L + Out[96]=.9 6 è!!! π H Erfi@D + Erfi@DL We can see that the results are identical to those arrived at using the single integral above.

4 CAS Exercise Examples for Chapter 5 à Section 5.3 Double Integrals in Polar Form ê5!!!!!!!! Example: Given the double integral Ÿ Ÿ ê5x x + dx d, complete each of the following. (a) Plot the cartesian region of integration in the x-plane. (b) Change each boundar curve of the Cartesian region in part (a) to its polar representation b solving its Cartesian equation for r and q. (c) Using part (b), plot the polar region of integration in the rq-plane. (d) Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from our plot in part (c) and evaluate the polar integral using Mathematica. Part (a) Since x = ê 5 is equivalent to = 5 x, the following Plot command can be used to plot the cartesian region of integration. In[97]:= Clear@x,, r, θd PlotA85 x,, 5 x<, 9x, 5, =, PlotRange > 8, <E; Part (b) Replacing with r sinq and x with r cosq, the Solve command can then be used to find the values of r and q. (Ignore the warning messages.)

5 CAS Exercise Examples for Chapter 5 5 In[99]:= Solve@r Cos@θD == r Sin@θD ê 5, θd Solve@r Cos@θD == r Sin@θD ê 5, θd Solve@r Sin@θD ==, rd Solve::ifun : Inverse functions are being used b Solve, so some solutions ma not be found; use Reduce for complete solution information. More Out[99]= 99θ ArcCosA E=, 9θ ArcCosA 6 6 E== Solve::ifun : Inverse functions are being used b Solve, so some solutions ma not be found; use Reduce for complete solution information. More Out[]= 99θ ArcCosA E=, 9θ ArcCosA 6 6 E== Out[]= 88r Csc@θD<< Part (c): The command PolarPlot is contained in the Graphics package and the package ComplexMap contains the command PolarMap used to plot a portion of the polar coordinate sstem. In[]:= << Graphics`Graphics` << Graphics`ComplexMap` The following command will plot the polar coordinate sstem for r = to.5. In[]:= PolarMap@Identit, 8,.5<D; Since our region of integration is bounded b the lines q =ArcCosA E and q=arccosa E, the next command is used to plot the portion of the polar coordinate sstem of interest. 6 6

6 6 CAS Exercise Examples for Chapter 5 In[5]:= polargr = PolarMapAIdentit, 8,.,.<, 9 ArcCosA E, ArcCosA E, ArcCosA E ArcCosA E =, AspectRatio >., Ticks > None, GridLines > NoneE; The upper bound on the region of integration is r = csc q and the corresponding graph of this function can be found using the PolarPlot command.

7 CAS Exercise Examples for Chapter 5 7 In[6]:= upperbd = PolarPlotACsc@θD, 9θ, ArcCosA E, ArcCosA E=, 6 6 PlotRange 88.5,.5<, 8,.<<, AspectRatio > E; The region of integration can now be displaed in the polar coordinate sstem.

8 8 CAS Exercise Examples for Chapter 5 In[7]:= Show@8upperbd, polargr<, PlotRange > 8, <D; Part (d) The integral is now converted to polar form and evaluated. This computation ma take longer in some cases than in others. ArcCosA E 6 In[8]:= ArcCosA E 6 Out[8]= 3 i k j CscA CscA 8 LogACosA 8 LogASinA SecA SecA Sin@θD r Cos@θD r θ Jπ ArcTanA 5 ENE + CscA Jπ+ ArcTanA 5 ENE CscA Jπ ArcTanA 5 Jπ ArcTanA 5 Jπ ArcTanA 5 ENE SecA Jπ+ ArcTanA 5 ENE + SecA Now the numerical approximation of the integral is obtained. In[9]:= N@%D Out[9]=.7938 Jπ ArcTanA 5 ENE + Jπ+ ArcTanA 5 ENE ENEE + 8 LogACosA ENEE 8 LogASinA Jπ+ ArcTanA 5 ENEE + Jπ+ ArcTanA 5 ENEE + Jπ ArcTanA 5 ENE Jπ+ ArcTanA 5 ENE z { In order to check the answer, suppose ou attempt to integrate in Cartesian coordinates.

9 CAS Exercise Examples for Chapter 5 9 In[]:= Out[]= 5 5 x!!!!!!!! x + x ArcCsch@5D 5 Despite the difference in form, we will see that this result agrees with that arrived at with polar coordinates. In[]:= N@%D Out[]=.7938 à Section 5. Triple Integrals in Rectangular Coordinates Evaluating triple integrals with Mathematica is completel analogous to computing double integrals. For example, the triple integral found in Example 3 in our textbook, is evaluated below. x In[]:= x z x 6 Out[]= For Exercises 9-5, some of the bounds might be easier to consider using polar coordinates. For example, in 9, the side bounds are simpl r =. The following code defines the functions, specifies the switch to polar coordinates for the function, then evaluates the triple integral in two forms. Here, the order of integration is immaterial, since all the bounds are constants using polar coordinates. In[3]:= f:= x z topolar = 8x r Cos@tD, r Sin@tD<; fp = f ê. topolar êê Simplif π rfp r t z Integrate@r fp,8t,, π<, 8r,, <, 8z,, <D N@%D Out[5]= r z Cos@tD Sin@tD Out[6]= Out[7]= π 8 π 8 Out[8]=.6598 We can compare this result to what we would have gotten without the polar coordinate switch. In[9]:= NIntegrate@f, 8x,, <, 8, Sqrt@ x^d, Sqrt@ x^d<, 8z,, <D Out[9]=.6598

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