Question

Use Green's Theorem to calculate the work done by the force \(F\) on a particle that is moving counterclockwise around the closed path \(C\). $$ \mathbf{F}(x, y)=x y \mathbf{i}+(x+y) \mathbf{j}, C: x^{2}+y^{2}=4 $$

Short answer

The work done by the force on the particle is \(4π\).

Step-by-step solution

Compute the components of F

The given force field is \(\mathbf{F}(x, y) = xy\mathbf{i} + (x+y)\mathbf{j}\). So, the components are \(M = xy\) and \(N = x+y\).

Compute the partial derivatives

By the definition of Green's theorem, we need to calculate \(\frac{∂N}{∂x} - \frac{∂M}{∂y}\). For our force field, \(\frac{∂N}{∂x} = 1\) and \(\frac{∂M}{∂y} = x\). Hence, \(\frac{∂N}{∂x} - \frac{∂M}{∂y} = 1 - x\).

Setup the double integral

Using polar coordinates, \(x=r\cos(θ)\) and \(y=r\sin(θ)\), we have \(\frac{∂N}{∂x} - \frac{∂M}{∂y} = 1 - r\cos(θ)\). The integral is over the region \(D\) inside the closed curve \(C\), which is a circle of radius 2. Thus, the double integral becomes \(\int_0^{2π} \int_0^2 (1 - r\cos(θ)) r dr dθ\).

Evaluate the inner integral

Taking integral over \(r\) first, we have \(\int_0^{2}(1 - r\cos(θ)) r dr = \frac{r^2}{2} - \frac{r^3\cos(θ)}{3}\Big|_0^{2} = 2 - \frac{8\cos(θ)}{3}\).

Evaluate the outer integral

Now, taking integral over \(θ\), we have \(\int_0^{2π} (2 - \frac{8\cos(θ)}{3}) dθ = 2θ - \frac{8\sin(θ)}{3}\Big|_0^{2π} = 4π\).

Apply Green's theorem

By Green's theorem, the line integral around the curve \(C\), which gives the work done by the force field on the particle moving along \(C\), equals the double integral over the region \(D\), so it is equal to \(4π\).