NAME: YOUR NAME GOES HEREMath 2130: Maple Homework #3Due: Tuesday, November 15thDo not email me your answers. I want a paper copy.Please use "ctrl+delete" or "command+delete" to delete unused commands & command prompts. Clean your homework up before printing!Initialization
Don't forget to evaluate these commands or the rest of the sheet will not work properly. Before you print out your homework, click on the triangle to collapse this section back down (so it doesn't print).restart;
with(plots):
with(plottools):
with(LinearAlgebra):
with(VectorCalculus):
#
# doubleRiemann computes Riemann sums for double integrals.
#
# f is a function Ex: f might be "(x,y) -> x^2 + y^2;"
# X and Y are variables with range Ex: X might be "x=0..4"
# The net volume under f restricted to the rectangle defined
# by X and Y's ranges is approximated using m * n rectangles
# (where m and n are positive integers).
#
# EvalType is a list describing the evaluation scheme. The
# default setting is midpoint rule. Ex: EvalType could be
# something like "[Left,Right]" in this case rectangles would
# be evaluated in their upper left hand corners.
# OutputType can be "Value" "RiemannSum" or "Plot". "Value"
# causes the function to return the riemann sum's value.
# "RiemannSum" causes the function to return an unevaluated
# double sum. "Plot" returns a plot of the rectangular boxes
# and the surface. The plot title states the exact and approximate
# values of the double integral.
# T and Tp are transparency settings for the boxes and graph
# respectively. 0 means opaque and 1 means totally transparent.
#
doubleRiemann := proc(f,X,Y,m,n,EvalType := [Midpoint,Midpoint],OutputType := Plot, T := 0, Tp := 0.95)
local i,j,a,b,c,d,xV,yV,Dx,Dy,xS,yS,x,y,rSum,approxVal,actualVal,pTitle,blocks,graph,points;
# grab variable names and range data: "x=a..b & y=c..d"
x := lhs(X); y := lhs(Y);
a := lhs(rhs(X)); b := rhs(rhs(X));
c := lhs(rhs(Y)); d := rhs(rhs(Y));
# compute delta x and delta y (lengths and widths of subrectangles).
Dx := (b-a)/m; Dy := (d-c)/n;
# x & y coords of box bases.
xV := [seq(a+i*Dx,i=0..m-1)];
yV := [seq(c+j*Dy,j=0..n-1)];
# Set delta x according to mid, right or left hand rule
if EvalType[1] = Midpoint then
xS := Dx/2;
elif EvalType[1] = Right then
xS := Dx;
elif EvalType[1] = Left then
xS := 0;
else
RETURN(cat("Invalid evaluation type: ",EvalType[1]));
end if:
# Set delta y according to mid, right or left hand rule
if EvalType[2] = Midpoint then
yS := Dy/2;
elif EvalType[2] = Right then
yS := Dy;
elif EvalType[2] = Left then
yS := 0;
else
RETURN(cat("Invalid evaluation type: ",EvalType[2]));
end if:
# return an unevaluated Riemann sum
rSum := Sum(Sum(f(a+i*Dx+xS,c+j*Dy+yS)*Dx*Dy,i=0..nops(xV)-1),j=0..nops(yV)-1);
if OutputType = RiemannSum then RETURN(rSum); end if:
# compute the Riemann sum
approxVal := sum(sum(evalf(f(xV[i]+xS,yV[j]+yS)*Dx*Dy),i=1..nops(xV)),j=1..nops(yV));
if OutputType = Value then RETURN(approxVal); end if;
# compute the actual value
actualVal := evalf(int(int(f(x,y),x=a..b),y=c..d));
# create the plot title
pTitle := cat("Actual value = ",convert(actualVal,string),"\134n Approximate Value = ",convert(approxVal,string));
# create plots of the boxes
blocks := {seq(seq(cuboid([xV[i],yV[j],0],[xV[i]+Dx,yV[j]+Dy,f(xV[i]+xS,yV[j]+yS)],color=gray,transparency=T),
i=1..nops(xV)),j=1..nops(yV))}:
# create plots of the evaluation points
points := {seq(seq(point([xV[i]+xS,yV[j]+yS,f(xV[i]+xS,yV[j]+yS)],symbol=diamond,symbolsize=10,color=red,transparency=T),
i=1..nops(xV)),j=1..nops(yV))}:
# create a plot of the graph itself
graph := plot3d(f(x,y),x=a..b,y=c..d,color=blue,transparency=Tp):
display(blocks,graph,points,title=pTitle);
end proc:#1 Use the "doubleRiemann" function defined in the initialization above to approximate the double integrals defined below.(a) Approximate the area under LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYoLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYmLUYsNiVRInhGJ0YvRjItSSNtb0dGJDYtUSIsRicvRjNRJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0YxLyUpc3RyZXRjaHlHRkUvJSpzeW1tZXRyaWNHRkUvJShsYXJnZW9wR0ZFLyUubW92YWJsZWxpbWl0c0dGRS8lJ2FjY2VudEdGRS8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHUSwwLjMzMzMzMzNlbUYnLUYsNiVRInlGJ0YvRjJGQUZBLUY+Ni1RIj1GJ0ZBRkMvRkdGRUZIRkpGTEZORlAvRlNRLDAuMjc3Nzc3OGVtRicvRlZGam4tRiM2JS1GLDYjUSFGJy1JJm1zcXJ0R0YkNiMtRiM2KC1JI21uR0YkNiRRIjFGJ0ZBLUY+Ni1RKCZtaW51cztGJ0ZBRkNGaG5GSEZKRkxGTkZQL0ZTUSwwLjIyMjIyMjJlbUYnL0ZWRl5wLUkmbWZyYWNHRiQ2KC1JJW1zdXBHRiQ2JUY6LUYjNiUtRmdvNiRRIjJGJ0ZBRi9GMi8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRictRiM2JS1GZ282JFEiOUYnRkFGL0YyLyUubGluZXRoaWNrbmVzc0dGaW8vJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGZ3EvJSliZXZlbGxlZEdGRUZqby1GYXA2KC1GZHA2JUZYRmZwRltxLUYjNiUtRmdvNiRRIjRGJ0ZBRi9GMkZjcUZlcUZocUZqcUZBRkFGXm9GQQ== over the rectangle 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 using "doubleRiemann" a grid of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbW5HRiQ2JFEiNkYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RKCZ0aW1lcztGJ0YvLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0Y4LyUpc3RyZXRjaHlHRjgvJSpzeW1tZXRyaWNHRjgvJShsYXJnZW9wR0Y4LyUubW92YWJsZWxpbWl0c0dGOC8lJ2FjY2VudEdGOC8lJ2xzcGFjZUdRLDAuMjIyMjIyMmVtRicvJSdyc3BhY2VHRkctRiw2JFEiM0YnRi9GLw== rectangles and lower lefthand rule. In addition to computing this approximation, make the doubleRiemann command output a plot. Use a second doubleRiemann command to output a double summation. Finally, use a couple of "int" commands to compute the exact answer (if you apply "evalf" to your exact answer, it should match what the doubleRiemann command gives as the actual value). f := (x,y) -> ????:
'f(x,y)' = f(x,y);doubleRiemann(f,????);doubleRiemann(f,???);(b) Consider the function 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 integrated over the rectangle 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. It is possible to choose a grid size and evaluation type for which doubleRiemann gives an exact answer. Find the smallest possible grid size and evaluation type (i.e. lower lefthand rule or lower midpoint rule, or...) which does this. g := (x,y) -> 1+2*x+3*y:
'g(x,y)' = g(x,y);doubleRiemann(g,???);Conclusions?If we use a ??? \303\227 ??? grid and ???? rule, the double Riemann sum gives an exact answer.Is there anything special about LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkjbWlHRiQ2JVEiZ0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYmLUYsNiVRInhGJ0YvRjItSSNtb0dGJDYtUSIsRicvRjNRJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0YxLyUpc3RyZXRjaHlHRkUvJSpzeW1tZXRyaWNHRkUvJShsYXJnZW9wR0ZFLyUubW92YWJsZWxpbWl0c0dGRS8lJ2FjY2VudEdGRS8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHUSwwLjMzMzMzMzNlbUYnLUYsNiVRInlGJ0YvRjJGQUZBRkE= which makes this possible? ????#2 Consider the region LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiRUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= bounded by 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, 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, and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYnLUkjbWlHRiQ2JVEieUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIj1GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EsMC4yNzc3Nzc4ZW1GJy8lJ3JzcGFjZUdGTC1GNjYtUSomdW1pbnVzMDtGJ0Y5RjtGPkZARkJGREZGRkgvRktRLDAuMjIyMjIyMmVtRicvRk5GUy1JI21uR0YkNiRRIjJGJ0Y5Rjk=.(a) Fix the bounds in the plot3d commands below so that only the surface of our region LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiRUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= is shown. [You may want to use "solve" command to determine some of these bounds.]I'll get you started with a few possibly relevant solve commands...solve({5-x^2=1+2*y},y);allvalues(solve({z=1+2*y,y=-2,z=5-x^2}));top := plot3d(5-x^2,x=-3..3,y=-3..3,color=red):
bottom := plot3d(1+2*y,x=-4..4,y=-3..3,color=green):
side := plot3d([x,-2,z],x=-4..4,z=-4..6,color=blue):
display({top,bottom,side},scaling=constrained,axes=boxed);(b) Compute the volume of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiRUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= using all six different orders of integration. [Of course, you should get the same answer each time.]int(int(int(1,z=???..???),y=???..???),x=???..???);int(int(int(1,z=???..???),x=???..???),y=???..???);int(int(int(1,y=???..???),z=???..???),x=???..???);int(int(int(1,y=???..???),x=???..???),z=???..???);int(int(int(1,x=???..???),z=???..???),y=???..???);int(int(int(1,x=???..???),y=???..???),z=???..???);(c) Find the centroid of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiRUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic=.m := ???;Myz := ???;Mxz := ???;Mxy := ???;Centroid := (1/m)*<Myz,Mxz,Mxy>;#3 Consider the solid region LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkjbWlHRiQ2JVEiRUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIn5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTEY5bounded by LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYqLUklbXN1cEdGJDYlLUkjbWlHRiQ2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYlLUkjbW5HRiQ2JFEiMkYnL0Y2USdub3JtYWxGJ0YyRjUvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnLUkjbW9HRiQ2LVEiK0YnRj4vJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRkkvJSlzdHJldGNoeUdGSS8lKnN5bW1ldHJpY0dGSS8lKGxhcmdlb3BHRkkvJS5tb3ZhYmxlbGltaXRzR0ZJLyUnYWNjZW50R0ZJLyUnbHNwYWNlR1EsMC4yMjIyMjIyZW1GJy8lJ3JzcGFjZUdGWC1JJm1mcmFjR0YkNigtRiw2JS1GLzYlUSJ5RidGMkY1RjhGQC1GIzYlLUY7NiRRIjRGJ0Y+RjJGNS8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGZ28vJSliZXZlbGxlZEdGSUZDLUZmbjYoLUYsNiUtRi82JVEiekYnRjJGNUY4RkAtRiM2JS1GOzYkUSI5RidGPkYyRjVGYm9GZW9GaG9Gam8tRkQ2LVEiPUYnRj5GR0ZKRkxGTkZQRlJGVC9GV1EsMC4yNzc3Nzc4ZW1GJy9GWkZccS1GOzYkRmRvRj5GPg==. (a) Define a modified spherical coordinate transform which simplifies the bounds for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiRUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= (sort of "ellipsoid" coordinates). Modify the plot of the sphere below so that it shows the plot of the ellipsoid which bounds LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiRUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic=. Then use Maple to compute the Jacobian matrix and Jacobian determinant of your transform. Please simplify your Jacobian determinant.plot3d([cos(theta)*sin(phi),sin(theta)*sin(phi),cos(phi)],theta=0..2*Pi,phi=0..Pi,scaling=constrained,axes=boxed);? Jacobiansimplify(???);(b) Find the volume of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiRUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic=. (c) Compute 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 . [Find the exact answer.]