How to rigorously treat integrals where both the integrand and the variable of integration are vectors and there's a dot product?

Question

First, I consider this expression to find the work done by the weight over an object close to Earth's surface: $$W_{over\ object}=\int_{\vec{y_{1}}}^{\vec{y_{2}}}\vec w \cdot d\vec y$$ $$W_{over\ object}=\int_{\vec{y_{1}}}^{\vec{y_{2}}}-mg \hat{\textbf{j}} \cdot d\vec y$$ I'm assuming both the force and displacement are in the same direction, downwards, and I develop the dot product and get the next expression: $$W_{over\ object}=\int_{y_{2}}^{y_{1}}mgdy$$ Because it's going downwards ($y_{2}<y_{1}$), I thought it was a good idea to switch the integration limits since the expression is a scalar now, so that the upper limit is the highest value for the integration variable. Then I get this: $$W_{over\ object}=mgy_{1}-mgy_{2}$$

Because of Newton's third law, I may find the work done by the object over the thing that does work over the object, which is Earth (I guess): $$W_{over\ Earth}=\int_{\vec{y_{1}}}^{\vec{y_{2}}}-\vec w \cdot d\vec y$$ $$W_{over\ Earth}=\int_{\vec{y_{1}}}^{\vec{y_{2}}}+mg \hat{\textbf{j}} \cdot d\vec y$$ Again, I assume displacement is downwards, thus $y_{2}<y_{1}$: $$W_{over\ Earth}=\int_{y_{2}}^{y_{1}}-mgdy$$ $$W_{over\ Earth}=-mgy_{1}+mgy_{2}$$ If we let $U_{1}=mgy_{1}$ and $U_{2}=mgy_{2}$, we have: $$W_{over\ object}=-W_{over\ Earth}=U_{2}-U_{1}=\Delta U$$

In order to make this scenario more general: $$F_{over\ object}dx=W_{over\ object}$$ $$-F_{over\ object}dx=-W_{over\ object}=dU$$ $$F_{over\ object}=-\frac{dU}{dx}$$ Changing notation a bit: $$F(x)=-\frac{dU(x)}{dx}$$

After all this, what I've done seems to work, but I don't think it's rigorous at all. How can I approach this kind of problem more appropriately? More specifically, how can I treat an integral where the integrand is a vectorial function (namely, a force function), and the variable of integration is a vector as well? I think the core of my problem is treating the dot product itself when integrating, in order to find the work done by a conservative force function over a displacement interval and its associated potential energy function, with a vectorial approach.

In this sense, I have doubts regarding the derivation of the potential energy function for this particular case and how to obtain it by using a more mathematically rigorous approach.

Answer 1

I'm assuming both the force and displacement are in the same direction . . . .

Do not make this assumption!

Let $\hat z$ be a unit vector pointing upwards.

The work done by an external force $\vec F(z) = F(z)\, \hat z$ in moving a body from position $z_1 \hat z$ to position $z_2 \hat z$ is

$\displaystyle \int ^{z_2} _{z_1} \vec F(z)\cdot d\vec z$.

Now the next step is where a significant error is often made by trying to decide the direction of the incremental displacement $d\vec z$ relative to the force $\vec F(\vec z)$.

Do not.

Let $\vec z = dz\,\hat z$ where $dz$ is the component of the incremental displacement in the $\hat z$ direction.

Thus, the work done by external force is $\displaystyle \int ^{z_2} _{z_1} F(z)\,\hat z\cdot dz\,\hat z = \int ^{z_2} _{z_1} F(z)\,dz$.

So what about the sign of $dz$?
That is taken care of by the limits of integration.

Let the component of the external force in the $\hat z$ direction be $+m\,g$, ie $\vec F(z) = +m\,g\,\hat z$, an external force pointing upwards.

After integration the work done by the external force is $mg(z_2-z_1)$.

If $z_2>z_1$ then the work done by the external force is positive and if $z_2<z_1$ then the work done by the external force is negative.

If you want to consider the work done by the gravitational force then the only thing that changes is that now $\vec F(z) = -m\,g\,\hat z$.