In this lab you will explore how Mathematica can be used to work with line integrals.
Consider the problem of a thin wire shaped like the spring,
, with density given by
. We can use what we know about mass and moments to find the
center of mass of the spring. First, input the parameterization,
x = 5 Cos[t];
y = 5 Sin[t];
z = t;
r1={x,y,z}
Next, we can get a picture,
ParametricPlot3D[r1,{t,0,8 Pi},BoxRatios -> {1,1,1},
AxesLabel -> {"x","y","z"}]
It looks like a four looped spring. Now we can plug in the
parameterization of the curve into the density function and find the
``ds'' for this curve,
density = x^2+y^2+z
dr1 = D[r1,t]
ds = Sqrt[dr1.dr1]
Here, we use the fact that mass = NIntegrate[density ds, {t,0,8 Pi}]
Mxy = NIntegrate[z density ds, {t,0,8 Pi}]
Mxz = NIntegrate[y density ds, {t,0,8 Pi}]
Myz = NIntegrate[x density ds, {t,0,8 Pi}]
where we let Mathematica do all the substitution. Now the center of
mass is given by,
centermass = {Myz/mass, Mxz/mass, Mxy/mass}
Useing the same curve as above, lets find the work done if the force is given by,
F=Simplify[{-x,-y,-z}/(Sqrt[x^2+y^2+z^2]^3)]
which is a radially inward pointing vector field, with the magnitude of the
vectors decaying as
We can do this in Mathematica as follows,
Fdotdr1=F.dr1
work1=NIntegrate[Fdotdr1,{t,0,8 Pi}]
Now let's try a straight line path between (5,0,0) and x=5;
y=0;
z=t;
r2={x,y,z}
F={-x,-y,-z}/(Sqrt[x^2+y^2+z^2]^3);
dr2 = D[r2,t];
Fdotdr2=F.dr2;
work2=NIntegrate[Fdotdr2,{t,0,8 Pi}]
In this case,
Useing a parameterization for the ellipse,
,
and the Green's Theorem form for finding area,
find the formula for the area of the ellipse.