Calculus IV - Lab 3

Vector Integration

Work all of the following integrals entirely within Mathematica (i.e., even if you see a cool shortcut proving the integral equals zero, do it anyway). You may not work any of this by "cutting and pasting" results from one place to another or working it out by hand.  (The one exception is if you need to complete the square...)  You must set up your work so that, if I were to make a change to the definition of the function or the region, I could go through and re-execute everything and it would work with the new definitions (well, I might have to change some limits of integration as well).  Be sure to give a brief discussion of orientation where applicable.  Also, for each problem, tell me whether you can use Green's Theorem, Stokes' Theorem, Gauss' Theorem, or none of these on it.  Briefly explain why or why not and, if it is possible, do so and compare your results.  Use Mathematica to graph the objects you are integrating over as well.

Useful tip:  You might find RegionPlot and RegionPlot3D useful in some of the graphing...  Another function you might find useful is Reduce (it can simplify inequalities).  I include a few brief examples below; you will want to experiment with them a bit.

RegionPlot[1≤(x^2)+3(y^2)≤4 && y≥x,{x,-3,3},{y,-3,3},Axes->True]

RegionPlot3D[0.5≤(x^2)+(y^2)+(z^2)≤1&&0.1z≤(x^2)+(y^2)≤z,{x,-2,2},{y,-2,2},{z,-2,2},PlotPoints->60]

Reduce[(x^2)+(y^2)≤1,Reals]

Notice, on Reduce, it sometimes gives you more results than you actually need.  Here, for example, you only need the middle set of inequalities.  (Remember that && is a logical "AND" and || is a logical "OR" statement.)  

I'll leave it to you to think about how these might be useful...

Problems:

(∫_( (OverVector[c]))) (x^3)((y^2)-(z^2))ds where (OverVector[C])(t)=⟨cos t,cos 2t sin t,sin 2t sin t⟩, 0≤t≤2π

This curve, aside from looking pretty cool, has an important property.  Identify this property and give me some reason to believe you.

(∫_( (OverVector[c]))) (x^2)dx+(y^3)z (e^x) dy+4(z^2) dz where (OverVector[C]) is composed of the line segments from (1,2,3) to (4,5,6) to (-6,3,1)

(∫_( (OverVector[c])))  (OverVector[F])(x,y,z)·(OverVector[ds]) where (OverVector[F])(x,y,z)=⟨y+z,-(x^2),-4(y^2)⟩ and (OverVector[C])(t) is the boundary of the parallelogram through the points:  (0,0,0), (1,2,3), (-3,6,9), and (-4,4,6)

∫(∫_S)  (x^2)z ln (y^2)dS where S is the surface of definted by (y^2)=(x^2)+(z^2), 0≤y≤4.

∫(∫_S) curl(OverVector[F])· (OverVector[dS]) where (OverVector[F])(x,y,z)=⟨(x^5),-(x^2)y,(x^5)(y^2)z⟩ and S is the portion of the cylinder (x^2)+(z^2)=1 that is between the planes x+y+z=0 and x+2y+z=2, oriented outwards.

∫(∫_S) (OverVector[F])· (OverVector[dS]) where (OverVector[F])(x,y,z)=⟨(x^2)+y,z sin x,(y^3)⟩ and S is the surface z=(x^2)+2(y^2) underneath the plane x+2z=1 (including the plane, so it is closed).