91Ó°ÊÓ

Compute the partial derivatives of the functions \(P\) and \(Q\) and then find the two-dimensional curl as follows: $$Q = x^{2}+y^{2} \Rightarrow \frac{\partial Q}{\partial x} = 2x$$ $$P = 0 \Rightarrow \frac{\partial P}{\partial y} = 0$$ The two-dimensional curl of the vector field \(\mathbf{F}\) is: $$Curl(\mathbf{F}) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 0 = 2x$$

Now, we need to parametrize the boundary curve of region \(R\). Since \(R\) is a unit circle, we can define the parametric equations as: $$x(\theta) = \cos\theta$$ $$y(\theta) = \sin\theta$$ where \(0 \leq \theta \leq 2\pi\) Calculate derivatives of the parametric equations: $$\frac{dx}{d \theta} = \frac{\partial x}{\partial\theta}=-\sin\theta$$ $$\frac{dy}{d \theta} = \frac{\partial y}{\partial\theta}=\cos\theta$$ Substitute the parametric equations into the vector field: $$\mathbf{F}(x(\theta), y(\theta))= \langle 0, \cos^2\theta + \sin^2\theta \rangle =\langle 0, 1 \rangle$$ So, the line integral is: $$\oint_{C} P dx + Q dy = \int_{0}^{2\pi}\langle 0, 1 \rangle \cdot (-\sin\theta, \cos\theta) d\theta = \int_{0}^{2\pi}\cos\theta d \theta $$ Evaluate the integral: $$\int_{0}^{2\pi}\cos\theta d \theta = \left[\sin\theta\right]_0^{2\pi} = 0$$

The double integral of the two-dimensional curl over region \(R\) can be computed as: $$\iint_{R} Curl(\mathbf{F}) dA = \iint_{R} 2x dA$$ To evaluate this, we will use polar coordinates. In polar coordinates, \(dA = r dr d\theta\), and \(x = r \cos\theta\). So, the integral becomes: $$\iint_{R} 2x dA = \int_{0}^{2\pi} \int_{0}^{1} 2r\cos\theta r dr d\theta$$ Evaluate the integral: $$\int_{0}^{2\pi} \int_{0}^{1} 2r^2\cos\theta dr d\theta = \left(\int_{0}^{2\pi} \cos\theta d\theta\right)\left(\int_{0}^{1} 2r^2 dr\right) = \left[\sin\theta\right]_0^{2\pi} \left[\frac{2}{3}r^3\right]_0^1$$ $$= 0 \cdot \left(\frac{2}{3}\right) = 0$$

We have calculated the line integral and the double integral as follows: $$\oint_{C} P dx + Q dy = 0$$ $$\iint_{R} Curl(\mathbf{F}) dA = 0$$ Since both integrals have the same value, we can conclude that the results are consistent, and Green's Theorem holds true for this vector field and region.