compute $\int_\gamma (\sin xy+xy\cos xy+2x )\ dx\ +(x^2\cos xy+e^y) \ dy=e-1$

Question

The question is : $$ F=(\sin xy+xy\cos xy+2x, \ x^2\cos xy+e^y)$$ Find a curve $\gamma$ such that $ \int_\gamma F=e-1$

I have tried to solve it like this: $$ \text{The field is conservative,}\quad F=\nabla f, \quad f=x\sin xy+e^y+x^2+C $$ I couldn't solve it fully I've just found the potential $f$

Any suggestion would be great, thanks

Answer 1

As you already found $f (x, y) = x\sin xy+e^y+x^2 + C$

Given the line integral is $e - 1$, it is easy to see that we can assign end point as $y = 1$ and starting point as $y = 0$.

Now to make sure other terms are zero, we can assign $x = 0$ for both the start and end points.

So starting points and end points are $A(0,0)$ and $B(0,1)$ respectively.

Now given that the line integral is path independent for the given vector field, it could be any smooth curve from point $A$ to $B$ - for example, a straight line or say, a semicircle of radius $\frac{1}{2}$ with center at $(0, \frac{1}{2})$.

Answer 2

Once you have that the field is conservative, you just need to find two points where the underlying scalar field assumes values of $e$ and $1$. Example points are $(x,y)=(0,1)$ and $(0,0)$. Thus $\gamma$ can be the line segment from $(0,1)$ to $(0,0)$.

Answer 3

You just need to find 2 points $(x_1,y_1)$, $(x_2,y_2)$ such that $f(x_2,y_2)-f(x_1,y_1) = e-1$. After you do this, just choose any $\gamma$ with those endpoints.

Answer 4

Let $ \gamma(t):=t(0,1)$ for $t \in [0,1].$ Then

$$\gamma(0)=(0,0)$$

and

$$ \gamma(1)=(0,1).$$

Then

$$\int_\gamma F=f( \gamma(1))-f(\gamma(0))=e-1.$$