Section 16.4: Green’s Theorem
MAT329, Fall, 2015
Andy Long

Clear[x,y,t];

One more from 3.3: #35, p. 1107: so what’s the function whose gradient works (mostly -- but not at the origin)?

f[x_,y_]:=ArcTan[y/x]
Simplify[D[f[x,y],x]]
Simplify[D[f[x,y],y]]

#1, p. 1113

x[t_]:=2Cos[t]
y[t_]:=2Sin[t]
r[t_]:={x[t],y[t]}
Q[x_,y_]:=x+y
P[x_,y_]:=x-y
F[x_,y_]:={P[x,y],Q[x,y]}
Integrate[Dot[F[x[t],y[t]],r'[t]],{t,0,2Pi}]
D[Q[x,y],x]
D[P[x,y],y]

#3, p. 1113

Clear[a,b,x,y]
x[t_]:=a Cos[t]
y[t_]:=b Sin[t]
Integrate[x[t]*y'[t],{t,0,2Pi}]

#19, p. 1114

x[t_]:=t-Sin[t]
y[t_]:=1-Cos[t]
(* If we take as our curve the origin to (2Pi,0), then over the top of the cycloid back to the origin, then
 the integral along the x-axis is zero, leaving 
	the integral along the cycloid, from right to left: *)
Integrate[x[t]*y'[t],{t,2Pi,0}]

#23, p. 1114

Clear[A,a]
A=Pi a^2/4
x[t_]:=a Cos[t]
y[t_]:=a Sin[t]
xbar=1/(2A)Integrate[x[t]^2*y'[t],{t,0,Pi/2}]
(* by symmetry, ybar=xbar! *)