1. Consider the vector field $\vec{F}=\langle x+y, y+z, x+z\rangle$ and the open surface $\mathcal{S}$ defined by $y=10-x^{2}-z^{2}$ for $y \geq 1$. Find the line integral $\oint_{\partial S} \vec{F} \bullet d \vec{r}$ in the following ways. (a) Parameterize the curve $\mathcal{C}=\partial \mathcal{S}$ as counter clockwise if viewed from $+\infty$ on the $y$-axis and solve the integral directly. (b) Apply Stokes' Theorem to the surface $\mathcal{S}$ oriented positive outward. (c) Apply Stokes' Theorem to the disk contained within $\mathcal{C}$ oriented in a consistent manner to the rest of the question.

Step 1 ：Parameterize the curve \(\mathcal{C}=\partial \mathcal{S}\) as counter clockwise if viewed from \(+\infty\) on the \(y\)-axis. The boundary of the surface \(\mathcal{S}\) is a circle in the plane \(y=1\), with radius \(\sqrt{10-1}=3\). We can parameterize this circle as \(\vec{r}(t) = \langle 3\cos(t), 1, 3\sin(t)\rangle\) for \(0 \leq t \leq 2\pi\).

Step 2 ：Compute the derivative of \(\vec{r}(t)\) with respect to \(t\), which is \(\vec{r}'(t) = \langle -3\sin(t), 0, 3\cos(t)\rangle\).

Step 3 ：Evaluate the vector field \(\vec{F}\) at \(\vec{r}(t)\), which is \(\vec{F}(\vec{r}(t)) = \langle 3\cos(t)+1, 1+3\sin(t), 3\cos(t)+3\sin(t)\rangle\).

Step 4 ：Compute the dot product of \(\vec{F}(\vec{r}(t))\) and \(\vec{r}'(t)\), which is \(\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) = 9\cos^2(t)\).

Step 5 ：Evaluate the line integral of \(\vec{F}\) over \(\mathcal{C}\), which is \(\oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} = \int_0^{2\pi} 9\cos^2(t) dt = 9\pi\).

Step 6 ：Apply Stokes' Theorem to the surface \(\mathcal{S}\) oriented positive outward. Stokes' Theorem states that \(\oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} = \iint_{\mathcal{S}} (\nabla \times \vec{F}) \cdot d\vec{S}\), where \(\nabla \times \vec{F}\) is the curl of \(\vec{F}\) and \(d\vec{S}\) is the outward-pointing surface element.

Step 7 ：Compute the curl of \(\vec{F}\), which is \(\nabla \times \vec{F} = \langle 1, -1, 0\rangle\).

Step 8 ：Parameterize the surface \(\mathcal{S}\) as \(\vec{S}(x,z) = \langle x, 10-x^2-z^2, z\rangle\) for \(x^2+z^2 \leq 9\). The outward-pointing surface element is \(d\vec{S} = \langle -2x, 1, -2z\rangle dx dz\).

Step 9 ：Compute the dot product of \(\nabla \times \vec{F}\) and \(d\vec{S}\), which is \((\nabla \times \vec{F}) \cdot d\vec{S} = -2x - 1\).

Step 10 ：Evaluate the double integral of this over \(\mathcal{S}\), which is \(\iint_{\mathcal{S}} (\nabla \times \vec{F}) \cdot d\vec{S} = 9\pi\).

Step 11 ：Apply Stokes' Theorem to the disk contained within \(\mathcal{C}\) oriented in a consistent manner to the rest of the question. The disk contained within \(\mathcal{C}\) is the disk in the plane \(y=1\) with radius 3. We can parameterize this disk as \(\vec{D}(r,\theta) = \langle r\cos(\theta), 1, r\sin(\theta)\rangle\) for \(0 \leq r \leq 3\) and \(0 \leq \theta \leq 2\pi\). The outward-pointing surface element is \(d\vec{D} = \langle \cos(\theta), 0, \sin(\theta)\rangle r dr d\theta\).

Step 12 ：Compute the dot product of \(\nabla \times \vec{F}\) and \(d\vec{D}\), which is \((\nabla \times \vec{F}) \cdot d\vec{D} = \cos(\theta)\).

Step 13 ：Evaluate the double integral of this over the disk, which is \(\iint_{\mathcal{D}} (\nabla \times \vec{F}) \cdot d\vec{D} = 0\).

Step 14 ：So, the line integral of \(\vec{F}\) over \(\mathcal{C}\) is \(\boxed{9\pi}\) by direct computation, \(\boxed{9\pi}\) by applying Stokes' Theorem to the surface \(\mathcal{S}\), and \(\boxed{0}\) by applying Stokes' Theorem to the disk contained within \(\mathcal{C}\). This is a contradiction, so there must be a mistake in the problem or in our computations.