1. **Problem Statement:** Evaluate the line integral $$\oint_C (2x^2 - y^2)\,dx + (x^2 + y^2)\,dy$$ where $C$ is the boundary of the region enclosed by the $x$-axis and the semicircle $x^2 + y^2 = 1$ in the upper half-plane. 2. **Green's Theorem:** Green's theorem relates a line integral around a simple closed curve $C$ to a double integral over the region $D$ enclosed by $C$: $$\oint_C P\,dx + Q\,dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$$ where $P = 2x^2 - y^2$ and $Q = x^2 + y^2$. 3. **Calculate partial derivatives:** $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2) = 2x$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2x^2 - y^2) = -2y$$ 4. **Substitute into Green's theorem integrand:** $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - (-2y) = 2x + 2y = 2(x + y)$$ 5. **Set up the double integral over $D$:** $D$ is the region bounded by the semicircle $x^2 + y^2 = 1$ and the $x$-axis in the upper half-plane, so $y \geq 0$. Using polar coordinates: $$x = r\cos\theta, \quad y = r\sin\theta$$ with $r$ from $0$ to $1$ and $\theta$ from $0$ to $\pi$. The area element is: $$dA = r\,dr\,d\theta$$ 6. **Rewrite the integrand in polar coordinates:** $$2(x + y) = 2(r\cos\theta + r\sin\theta) = 2r(\cos\theta + \sin\theta)$$ 7. **Integral becomes:** $$\iint_D 2(x + y) dA = \int_0^{\pi} \int_0^1 2r(\cos\theta + \sin\theta) \cdot r \, dr \, d\theta = \int_0^{\pi} (\cos\theta + \sin\theta) \int_0^1 2r^2 \, dr \, d\theta$$ 8. **Evaluate inner integral:** $$\int_0^1 2r^2 \, dr = 2 \cdot \frac{r^3}{3} \Big|_0^1 = \frac{2}{3}$$ 9. **Integral reduces to:** $$\frac{2}{3} \int_0^{\pi} (\cos\theta + \sin\theta) \, d\theta = \frac{2}{3} \left( \int_0^{\pi} \cos\theta \, d\theta + \int_0^{\pi} \sin\theta \, d\theta \right)$$ 10. **Evaluate integrals:** $$\int_0^{\pi} \cos\theta \, d\theta = \sin\theta \Big|_0^{\pi} = 0 - 0 = 0$$ $$\int_0^{\pi} \sin\theta \, d\theta = -\cos\theta \Big|_0^{\pi} = -(-1) - (-1) = 2$$ 11. **Final value:** $$\frac{2}{3} (0 + 2) = \frac{4}{3}$$ **Answer:** The value of the line integral is $$\boxed{\frac{4}{3}}$$.