Green's Theorem Line Integral

To evaluate the line integral $\oint_C P \, dx + Q \, dy$ where $P = x - y^3 - y$ and $Q = x^3 + y^2$, and $C$ is the unit circle $x^2 + y^2 = 1$ traversed counter-clockwise, we can use Green's Theorem.

Green's Theorem states:

$$\oint_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$$

where $D$ is the region enclosed by $C$.

1. Compute the Partial Derivatives:

   $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^3 + y^2) = 3x^2$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x - y^3 - y) = -3y^2 - 1$$

   Thus,

   $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3x^2 + 3y^2 + 1$$

2. Set Up the Double Integral:

   The region $D$ is the unit disk $x^2 + y^2 \leq 1$. Switching to polar coordinates where $x = r\cos\theta$ and $y = r\sin\theta$, the integrand becomes:

   $$3x^2 + 3y^2 + 1 = 3r^2 + 1$$

   The area element $dA$ in polar coordinates is $r \, dr \, d\theta$.

   So, the integral becomes:

   $$\iint_D (3r^2 + 1) \, r \, dr \, d\theta = \int_0^{2\pi} \int_0^1 (3r^3 + r) \, dr \, d\theta$$

3. Evaluate the Integral:

   • Integrate with respect to $r$:

     $$\int_0^1 (3r^3 + r) \, dr = \left[ \frac{3r^4}{4} + \frac{r^2}{2} \right]_0^1 = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$$

   • Integrate with respect to $\theta$:

     $$\int_0^{2\pi} \frac{5}{4} \, d\theta = \frac{5}{4} \times 2\pi = \frac{5}{2}\pi$$

Final Answer:

$$\oint_C P \, dx + Q \, dy = \frac{5}{2} \pi$$

Answer:
After simplifying, the integral is five-halves times π. Thus,

∮C P dx + Q dy = (5⁄2) π