In[7]:=
D[y + Cos[x], y] == D[x-1, x]
Out[7]=
True
b) Evaluate the integral of F . dx along C_1, where
where C_1 is the closed loop given by x(t) = cos(t),
y = t^2 - 2 pi t, 0 <= t <= 2 pi.
Because the vector field is exact, the integral around any closed loop is 0,
It is not necessary to parametrize.
c) Evaluate the integral of F . dx,
where C_2 is the non-closed curve given by x(t) = cos(t), y = t^2 - 2 pi t,
0 <= t <= pi
Because the vector field is exact, the vector field is the gradient of a scalar
function f[x,y], and the integral is just f[end point] - f[ starting point]. I
will solve this by finding the scalar function f. Another valid possibility
is to use the path-independence property, and do the integral directly
along a straight linefrom (1,0) to (-1,- pi^2).
To find x, integrate the components of the vector field and compare terms
to fix the "constants" of integration, by which I mean the arbitrary
functions of one variable alone . I will write them explicitly:
In[8]:=
f[x, y] = Integrate[y + Cos[x], x] + C1[y]
Out[8]=
x y + C1[y] + Sin[x]
In[9]:=
f[x, y] = Integrate[x - 1, y] + C2[x]
Out[9]=
(-1 + x) y + C2[x]
In[10]:=
% == %%
Out[10]=
(-1 + x) y + C2[x] == x y + C1[y] + Sin[x]
Evidently, we can take C1[y] = -y and C2[x] = Sin[x], so
In[11]:=
Clear[f]
In[12]:=
f[x_,y_] = x y + Sin[x] - y
Out[12]=
-y + x y + Sin[x]
Checking:
In[13]:=
{D[f[x,y], x], D[f[x,y], y]}
Out[13]=
{y + Cos[x], -1 + x}
The value of the integral is
In[14]:=
(%% /.{x -> -1, y -> - Pi^2}) - (%% /.{x -> 1, y -> 0})
Out[14]=
2 2 Pi - 2 Sin[1]
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