91Ó°ÊÓ

The two-dimensional divergence of a vector field \(\mathbf{F}=\langle P,Q \rangle\) is given by: $$\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$$ For the given vector field \(\mathbf{F}=\langle x^{2}+y^{2}, 0\rangle\), we have \(P=x^{2}+y^{2}\) and \(Q=0\). Now we will find their partial derivatives with respect to \(x\) and \(y\) respectively. $$\frac{\partial P}{\partial x} = \frac{\partial (x^{2}+y^{2})}{\partial x} = 2x$$ $$\frac{\partial Q}{\partial y} = \frac{\partial (0)}{\partial y} = 0$$ Now, let's compute the divergence of the vector field: $$\nabla \cdot \mathbf{F} = 2x + 0 = 2x$$

Green's Theorem (flux form) states that if \(\mathbf{F} = \langle P, Q\rangle\) is a continuously differentiable force field, the circulation around the closed curve \(C\) enclosing region \(R\) is equal to the integral of the two-dimensional divergence on the region \(R\): $$\oint_{C} \mathbf{F}\cdot \mathrm{d}\mathbf{s} = \int\int_{R}(\nabla \cdot \mathbf{F})\,\mathrm{d}A$$ In our case, the region \(R\) is the interior of a circle with radius 1, \(R = \{(x, y): x^2 + y^2 \leq 1\}\).

Let's parameterize the curve \(C\) in terms of polar coordinates \(x = r\cos\theta\) and \(y = r\sin\theta\), where \(r = 1\) and \(0\leq\theta\leq2\pi\). Then, the parameterized curve \(C\) is given by: $$\bold{r}(\theta) = \langle 1\cos\theta, 1\sin\theta\rangle$$ The tangent vector \(\bold{r}'(\theta)\) is: $$\bold{r}'(\theta) = \langle -1\sin\theta, 1\cos\theta\rangle$$ Next, we find \(\mathbf{F}( \bold{r}(\theta))\) as follows: $$\mathbf{F}( \bold{r}(\theta))=\langle (1\cos\theta)^{2}+(1\sin\theta)^{2}, 0\rangle=\langle 1, 0\rangle$$ Now, let's compute the dot product \(\mathbf{F}( \bold{r}(\theta))\cdot\bold{r}'(\theta)\): $$\mathbf{F}( \bold{r}(\theta))\cdot\bold{r}'(\theta)=\langle 1, 0\rangle\cdot\langle -1\sin\theta, 1\cos\theta\rangle= -1\sin\theta$$ Now, we can evaluate the line integral: $$\oint_{C} \mathbf{F}\cdot \mathrm{d}\mathbf{s} = \int_{0}^{2\pi}(-1\sin\theta)\,d\theta = [\cos\theta]_{0}^{2\pi} = (\cos2\pi - \cos0) = 0$$

Since \(\nabla\cdot\mathbf{F} = 2x\), we can write the double integral in polar coordinates: $$\int\int_{R}(2x)\,\mathrm{d}A = \int\int_{R}(2r\cos\theta)r\,\mathrm{d}r\,\mathrm{d}\theta = 2\int_{0}^{2\pi}\int_{0}^{1} r^{2}\cos\theta\,\mathrm{d}r\,\mathrm{d}\theta$$ Now, we can evaluate the double integral by first integrating with respect to \(r\): $$2\int_{0}^{2\pi}\left[\frac{1}{3}r^{3}\cos\theta\right]_{0}^{1}\,\mathrm{d}\theta = \frac{2}{3}\int_{0}^{2\pi}\cos\theta\,\mathrm{d}\theta$$ Then, integrate with respect to \(\theta\): $$\frac{2}{3}\left[\sin\theta\right]_{0}^{2\pi} = \frac{2}{3}(\sin{2\pi} - \sin{0}) = 0$$

We have found that both integrals are equal: $$\oint_{C} \mathbf{F}\cdot \mathrm{d}\mathbf{s} = 0$$ $$\int\int_{R}(\nabla \cdot \mathbf{F})\,\mathrm{d}A = 0$$ Since both integrals have the same value, this exercise is consistent with Green's Theorem.