MATH 149LABORATORY ASSIGNMENT 9GEOMETRIC APPLICATIONSOF INTEGRATION The definite integral can be used to solve a variety of problems from geometry; examples are finding areas between curves, lengths of curves, volumes and surface area of three dimensional solids, and centroids of plane regions. In this laboratory we will use MAPLE to calculate the area between curves and the volume of solids of revolution. The volume problem will also give us the opportunity to use MAPLE's 3D graphing abilities.AREA BETWEEN CURVES In class we showed that if f (x) > g (x) for all x in [ a , b ] then the area A of the region enclosed by the curves y = f (x), y = g (x), and the lines x = a a nd x = b is given byA = d⌠⌡⎮⎮ab − ()f x ()g xx . In practice, a large part of the problem in using this formula involves determining exactly where f (x) > g (x) and where g (x) > f (x). We illustrate with an example: EXAMPLE Find the total area of all regions enclosed by the curves f(x) = x + sin(2x) and g(x) = x3. The first thing to do is to plot the graphs:> with(plots):Warning, the name changecoords has been redefined > f:=x->x+sin(2*x); := f → x + x ()sin 2 x> g:=x->x^3;:= g → xx3> c:=textplot({[1.3,4,`g`],[3,1.5,`f`]}):> p:=plot({f,g},-Pi..Pi,-5..5):> > display({c,p});Note that the curves intersect at the origin, since f(0) = g (0) = 0. Next, we try find the X-coordinates of the other two points of intersection of the curves.> solve(f(x)=g(x),x);12()RootOf − − + 8()sin _Z 4 _Z _Z3Apparently there is no "nice" formula for the solutions, but we can still use "fsolve" to find decimal representations for them.> X[1]:=fsolve(f(x)=g(x),x=-2..-1); := X1-1.229835717> X[2]:=fsolve(f(x)=g(x),x,x=1..2); := X21.229835717Now we determine the interval or intervals where f(x) > g(x), and the interval or intervals where g(x) > f(x).From the graph, one can see that g (x) > f(x) for x in [ X1, 0 ] and that f (x) > g (x) for x in [ 0 , X2 ].Finally, we set up the integrals and evaluate them:> Area:=Int(g(x)-f(x),x=X[1]..0)+Int(f(x)-g(x),x=0..X[2]); := Area + d⌠⌡⎮⎮-1.2298357170 − − x3x ()sin 2 xx d⌠⌡⎮⎮01.229835717 + − x ()sin 2 xx3x> value(%);2.145037216 VOLUMES OF SOLIDS OF REVOLUTION A solid generated by revolving a curve y = f (x) about the X-axis is called a solid of revolution. Examples of such solids are cylinders, cones and spheres. The commandplot3d([x, f(x)*cos(t), f(x)*sin(t)], x=a..b, t=0..2*Pi);plots the surface generated by revolving y = f (x) , a < x < b about the X-axis. As an example, let f (x) = x2 , 0 < x < 1.> restart:> with(plots):Warning, the name changecoords has been redefined > f:=x->x^2; := f → xx2> plot3d([x, f(x)*cos(t), f(x)*sin(t)], x=0..1, t=0..2*Pi);Click on the figure, and use the toolbar to rotate the solid, and change the appearance of the coordinate system. You can also rotate the solid by moving the cursor while holding down the left mouse button.You may plot several graphs on the same coordinate system. For example, let's intersect the above graph with the X-Y plane.> A:=(%):> B:=plot3d([x,y,0],x=0..1.5,y=-1.5..1.5):> display3d({A,B});The formula for the volume V of the solid resulting from rotating y = f (x) , a < x < b , about the X-axis isV=π d⌠⌡⎮⎮ab()f x2xThus, in the above exampleV = π d⌠⌡⎮⎮01()x22x,and therefore> V:=int(Pi*x^4,x=0..1); := Vπ5> EXERCISES 1. Find the total area of all regions enclosed by the curves y = x2()cos x and y =− x3x 2. Make an appropriate choice of f (x) to generate graphs of: a) a sphere of radius 4, b) a cylinder of height 9 and radius 2, c) a cone with height 9 and base radius 2. Use MAPLE and the formula for the volume of a solid of revolution to find the volumes of the above solids. Show that your results are consistent with the usual formulasVolume(Sphere) = 4 π r33 , Volume(Cylinder) = πr2h Volume(Cone) = (1/3)πr2 3. The command animate3d([x, f(x)*cos(t*d), f(x)*sin(t*d)], x=a..b, t=0..2*Pi,d=0..1, axes=NORMAL); produces a "movie" of the generation of the surface. Try it. (you don't have to turn it in.) To get a less jerky movie, add "frames=20" after "d=0..1" in the
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