## 3. (8 points). Let F(x,y,z) := (y + cos(x))i + (x-1) j

a) Is this vector field exact ?

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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