Work Done by a Force Field When Given Integral With dx and dy

Question

This homework question is absolutely nothing like anything else my teacher taught us in our course. The problem is as follows:

Find the work done by the force field $\vec{F} (x,y) = y \ \vec{i} - x \ \vec{j}$ acting on an object as it moves along the parabola $y=x^2-1$ from $(1,0)$ to $(-2,3)$ is given by:

$\int_{C}^{}y \ dx-x \ dy$

I hate to ask a question like this without showing any work, but it is because I do not even have an idea of where to begin with this problem. We never discussed how to solve problems like these in class, and I'd greatly appreciate if someone could help me with this question.

Answer 1

The work done by the force field $\vec F(x,y)=y\ \hat i-x\ \hat j\ $ is given by $\vec F\cdot\vec {ds}$, where $\vec {ds}=dx\ \hat i+dy\ \hat j$, is the infinitesimal displacement of the particle.

$\thinspace \implies dW=\vec F\cdot\vec {ds}=(y\ \hat i-x\ \hat j)\cdot(dx\ \hat i+dy\ \hat j)=y\ dx-x\ dy\\\implies W=\int_C y\ dx-x\ dy$

The line integral is taken along $C$, the path that the particle takes. It is given to you that on the path, $y=x^2-1$. This means that on the path, the force field is $$\vec F(x,y)=\vec F(x,x^2-1)=(x^2-1)\ \hat i-x\ \hat j\ $$ and the infinitesimal displacement of the particle along the $y$-axis, $dy=d(x^2-1)=2x\ dx$

$$\implies W=\int_{x=1}^{x=-2}(x^2-1)dx-x\cdot2x\ dx\\ =-\int_{x=1}^{x=-2}(x^2+1)dx$$

Answer 2

Easily just by substituting $y=x^2-1$ in the integral we obtain$$W{=\int_{C}^{}y \ dx-x \ dy\\=\int_{C}^{}x^2-1 \ dx-x \ d(x^2-1)\\=\int_C x^2-1 \ dx-2x^2 \ dx\\=\int_C-x^2-1\ dx\\=-{x^3\over 3}-x\Bigg|_{x=1}^{x=-2}\\=6\text{ Joules}}$$