Evaluate The Surface Integral Of Vector Field F Over Plane S

In the realm of vector calculus, surface integrals play a crucial role in evaluating the flow of vector fields across surfaces. This article delves into the process of evaluating a surface integral, specifically focusing on the integral of a vector field ${\mathbf{F}}$ over a surface ${S}$ defined by a plane in the first octant. We will explore the fundamental concepts, methodologies, and step-by-step calculations involved in solving such problems. This exploration will not only enhance your understanding of vector calculus but also equip you with the skills to tackle similar problems with confidence.

The problem at hand involves evaluating the surface integral

${ \iint_S \mathbf{F} \cdot d\mathbf{S}, }$

where the vector field ${\mathbf{F}}$ is given by

${ \mathbf{F} = (x+y)\mathbf{i} + (2x - z)\mathbf{j} + (y+z)\mathbf{k} }$

and the surface ${S}$ is the portion of the plane

${ 3x + 2y + z = 6 }$

that lies in the first octant. The first octant is the region where all three coordinates, ${x}$, ${y}$, and ${z}$, are non-negative.

Before diving into the solution, it's essential to grasp the underlying concepts. A surface integral is a generalization of a double integral to integration over a surface. When we evaluate the surface integral of a vector field, we are essentially calculating the flux of the vector field across the surface. The flux represents the amount of the vector field that flows through the surface. In simpler terms, imagine the vector field as the velocity field of a fluid; the surface integral then measures the rate at which the fluid flows through the surface.

The vector field ${\mathbf{F}}$ in this problem is a function that assigns a vector to each point in space. The surface ${S}$ is a portion of a plane, which is a two-dimensional surface. The differential surface area element ${d\mathbf{S}}$ is a vector that is normal to the surface and has a magnitude equal to the infinitesimal area element ${dS}$. The dot product ${\mathbf{F} \cdot d\mathbf{S}}$ gives the component of ${\mathbf{F}}$ that is normal to the surface, which is what we need to compute the flux.

To evaluate the surface integral, we need to parameterize the surface ${S}$. This involves expressing the coordinates ${x}$, ${y}$, and ${z}$ in terms of two parameters, say ${u}$ and ${v}$. Once we have a parameterization, we can compute the normal vector to the surface. The surface integral then becomes a double integral over the parameter domain.

Here’s a step-by-step breakdown of the methodology:

1. Parameterize the surface ${S}$: Express the coordinates ${x}$, ${y}$, and ${z}$ in terms of two parameters, ${u}$ and ${v}$.
2. Compute the tangent vectors: Calculate the partial derivatives of the parameterization with respect to ${u}$ and ${v}$, denoted as ${\mathbf{r}_u}$ and ${\mathbf{r}_v}$, respectively.
3. Find the normal vector: Compute the cross product of the tangent vectors, ${\mathbf{n} = \mathbf{r}_u \times \mathbf{r}_v}$. This vector is normal to the surface.
4. Express the vector field in terms of the parameters: Substitute the parameterization into the vector field ${\mathbf{F}}$ to express it in terms of ${u}$ and ${v}$.
5. Compute the dot product: Calculate the dot product of the vector field and the normal vector, ${\mathbf{F} \cdot \mathbf{n}}$.
6. Set up the double integral: Determine the limits of integration for the parameters ${u}$ and ${v}$ based on the region of the surface ${S}$.
7. Evaluate the double integral: Compute the double integral of ${\mathbf{F} \cdot \mathbf{n}}$ over the parameter domain.

Let's apply this methodology to our problem. The surface ${S}$ is given by the equation

${ 3x + 2y + z = 6. }$

Since we are in the first octant, we have ${x \geq 0}$, ${y \geq 0}$, and ${z \geq 0}$. We can express ${z}$ in terms of ${x}$ and ${y}$ as

${ z = 6 - 3x - 2y. }$

1. Parameterize the Surface

We can parameterize the surface using ${x}$ and ${y}$ as parameters. Let ${x = u}$ and ${y = v}$. Then,

${ z = 6 - 3u - 2v. }$

The parameterization ${\mathbf{r}(u, v)}$ is given by

${ \mathbf{r}(u, v) = u\mathbf{i} + v\mathbf{j} + (6 - 3u - 2v)\mathbf{k}. }$

2. Compute the Tangent Vectors

Next, we compute the partial derivatives of ${\mathbf{r}}$ with respect to ${u}$ and ${v}$:

${ \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} = \mathbf{i} - 3\mathbf{k} }$

${ \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} = \mathbf{j} - 2\mathbf{k} }$

3. Find the Normal Vector

The normal vector ${\mathbf{n}}$ is the cross product of ${\mathbf{r}_u}$ and ${\mathbf{r}_v}$:

${ \mathbf{n} = \mathbf{r}_u \times \mathbf{r}_v = (\mathbf{i} - 3\mathbf{k}) \times (\mathbf{j} - 2\mathbf{k}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -3 \\ 0 & 1 & -2 \end{vmatrix} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k}. }$

This normal vector points upwards, which is consistent with the orientation of the surface in the first octant.

4. Express the Vector Field in Terms of the Parameters

Now, we express the vector field ${\mathbf{F}}$ in terms of ${u}$ and ${v}$. Recall that

${ \mathbf{F} = (x+y)\mathbf{i} + (2x - z)\mathbf{j} + (y+z)\mathbf{k}. }$

Substituting ${x = u}$, ${y = v}$, and ${z = 6 - 3u - 2v}$, we get

${ \mathbf{F}(u, v) = (u+v)\mathbf{i} + (2u - (6 - 3u - 2v))\mathbf{j} + (v + (6 - 3u - 2v))\mathbf{k} = (u+v)\mathbf{i} + (5u + 2v - 6)\mathbf{j} + (-3u - v + 6)\mathbf{k}. }$

5. Compute the Dot Product

We compute the dot product of ${\mathbf{F}(u, v)}$ and ${\mathbf{n}}$:

${ \mathbf{F} \cdot \mathbf{n} = ((u+v)\mathbf{i} + (5u + 2v - 6)\mathbf{j} + (-3u - v + 6)\mathbf{k}) \cdot (3\mathbf{i} + 2\mathbf{j} + \mathbf{k}) }$

${ = 3(u+v) + 2(5u + 2v - 6) + (-3u - v + 6) = 3u + 3v + 10u + 4v - 12 - 3u - v + 6 = 10u + 6v - 6. }$

6. Set Up the Double Integral

To set up the double integral, we need to determine the limits of integration for ${u}$ and ${v}$. The surface ${S}$ is in the first octant, so ${x \geq 0}$, ${y \geq 0}$, and ${z \geq 0}$. This gives us the following inequalities:

${ u \geq 0, \quad v \geq 0, \quad 6 - 3u - 2v \geq 0. }$

The last inequality can be rewritten as

${ 3u + 2v \leq 6. }$

When ${v = 0}$, we have ${3u \leq 6}$, so ${u \leq 2}$. When ${u = 0}$, we have ${2v \leq 6}$, so ${v \leq 3}$. Thus, the region of integration in the ${uv}$-plane is a triangle with vertices at ${(0, 0)}$, ${(2, 0)}$, and ${(0, 3)}$. We can describe this region as

${ 0 \leq u \leq 2, \quad 0 \leq v \leq 3 - \frac{3}{2}u. }$

The surface integral becomes the double integral

${ \iint_S \mathbf{F} \cdot d\mathbf{S} = \int_0^2 \int_0^{3 - \frac{3}{2}u} (10u + 6v - 6) \, dv \, du. }$

7. Evaluate the Double Integral

Finally, we evaluate the double integral:

${ \int_0^2 \int_0^{3 - \frac{3}{2}u} (10u + 6v - 6) \, dv \, du = \int_0^2 \left[ 10uv + 3v^2 - 6v \right]_0^{3 - \frac{3}{2}u} \, du }$

${ = \int_0^2 \left[ 10u(3 - \frac{3}{2}u) + 3(3 - \frac{3}{2}u)^2 - 6(3 - \frac{3}{2}u) \right] \, du }$

${ = \int_0^2 \left[ 30u - 15u^2 + 3(9 - 9u + \frac{9}{4}u^2) - 18 + 9u \right] \, du }$

${ = \int_0^2 \left[ 30u - 15u^2 + 27 - 27u + \frac{27}{4}u^2 - 18 + 9u \right] \, du }$

${ = \int_0^2 \left[ -\frac{33}{4}u^2 + 12u + 9 \right] \, du }$

${ = \left[ -\frac{11}{4}u^3 + 6u^2 + 9u \right]_0^2 = -\frac{11}{4}(8) + 6(4) + 9(2) = -22 + 24 + 18 = 20. }$

Therefore, the surface integral is:

${ \iint_S \mathbf{F} \cdot d\mathbf{S} = 20. }$

In this comprehensive guide, we have successfully evaluated the surface integral of the vector field ${\mathbf{F}}$ over the surface ${S}$, which is the portion of the plane ${3x + 2y + z = 6}$ in the first octant. We began by understanding the fundamental concepts of surface integrals and vector fields, followed by a detailed methodology. The step-by-step solution involved parameterizing the surface, computing tangent and normal vectors, expressing the vector field in terms of parameters, and finally, evaluating the double integral. This process not only provides the numerical answer but also deepens the understanding of how vector calculus is applied in practice. This article has equipped you with a solid foundation to tackle similar surface integral problems in the future. Remember, the key to mastering these concepts is practice and a thorough understanding of the underlying principles.

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