Let F(x,y,z) := (y + cos(x))i + (x-1) j

Answer. Yes, because

*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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