Surface Integral Evaluation Vector Field Over Plane In First Octant

Introduction

In vector calculus, surface integrals play a crucial role in evaluating the flux of a vector field across a given surface. This article delves into the process of evaluating a surface integral of a specific vector field F over a planar surface S situated in the first octant. Understanding this process is fundamental in various fields such as physics, engineering, and computer graphics, where vector fields and surface interactions are commonly encountered. This exploration will not only enhance your understanding of vector calculus but also equip you with the skills necessary to tackle similar problems in diverse applications.

Problem Statement

We are tasked with evaluating the surface integral denoted as ${ \iint_S \mathbf{F} \cdot d\mathbf{s} }$, where the vector field ${ \mathbf{F} }$ is defined as ${ \mathbf{F} = (x+y)\mathbf{i} + (2x - z)\mathbf{j} + (y+z)\mathbf{k} }$. The surface S is a portion of the plane described by the equation ${ 3x + 2y + z = 6 }$, specifically the part that lies within the first octant. The first octant is the region where all three coordinates, x, y, and z, are positive. Our goal is to compute the flux of the vector field ${ \mathbf{F} }$ across the surface S. To achieve this, we will parameterize the surface, compute the normal vector, and then evaluate the double integral.

Step 1: Parameterizing the Surface

To begin, we need to parameterize the surface S. Since S is a plane, we can express one of the variables in terms of the other two. From the equation ${ 3x + 2y + z = 6 }$, we can easily solve for z to get ${ z = 6 - 3x - 2y }$. This allows us to describe any point on the plane using x and y as parameters. We can define a vector function ${ \mathbf{r}(x, y) }$ as:

${ \mathbf{r}(x, y) = x\mathbf{i} + y\mathbf{j} + (6 - 3x - 2y)\mathbf{k} }$

This parameterization maps each pair of (x, y) values to a point on the plane. However, since we are only interested in the part of the plane in the first octant, we need to determine the bounds for x and y. The first octant is defined by ${ x \geq 0 }$, ${ y \geq 0 }$, and ${ z \geq 0 }$. Plugging ${ z = 6 - 3x - 2y }$ into the inequality ${ z \geq 0 }$, we get:

${ 6 - 3x - 2y \geq 0 }$

Rearranging this inequality gives:

${ 3x + 2y \leq 6 }$

This inequality, along with ${ x \geq 0 }$ and ${ y \geq 0 }$, defines the region in the xy-plane over which we will integrate. To find the bounds for x and y, we first consider the intercepts. When ${ y = 0 }$, we have ${ 3x = 6 }$, which gives ${ x = 2 }$. When ${ x = 0 }$, we have ${ 2y = 6 }$, which gives ${ y = 3 }$. Thus, the region of integration D in the xy-plane is a triangle bounded by the lines ${ x = 0 }$, ${ y = 0 }$, and ${ 3x + 2y = 6 }$. We can describe this region as:

${ D = \{(x, y) \mid 0 \leq x \leq 2, 0 \leq y \leq 3 - \frac{3}{2}x \} }$

This parameterization and the bounds for x and y set the stage for the next step: computing the normal vector.

Step 2: Computing the Normal Vector

The normal vector to the surface S is crucial for evaluating the surface integral. It provides the direction in which we are measuring the flux of the vector field. To find the normal vector, we compute the partial derivatives of the parameterization ${ \mathbf{r}(x, y) }$ with respect to x and y, and then take their cross product. The partial derivatives are:

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

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

Now, we compute the cross product:

${ \mathbf{n} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -3 \\ 0 & 1 & -2 \end{vmatrix} = (0 - (-3))\mathbf{i} - (1(-2) - 0)\mathbf{j} + (1 - 0)\mathbf{k} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} }$

This normal vector ${ \mathbf{n} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} }$ points in the direction away from the origin, which is consistent with the orientation of the surface as defined by the equation ${ 3x + 2y + z = 6 }$. This vector is essential for determining the orientation of the surface and ensuring we calculate the flux in the correct direction. With the normal vector in hand, we can now proceed to express the vector field ${ \mathbf{F} }$ in terms of the parameters x and y.

Step 3: Expressing the Vector Field in Terms of Parameters

To evaluate the surface integral, we need to express the vector field ${ \mathbf{F} }$ in terms of the parameters x and y. Recall that ${ \mathbf{F} = (x+y)\mathbf{i} + (2x - z)\mathbf{j} + (y+z)\mathbf{k} }$ and ${ z = 6 - 3x - 2y }$. Substituting the expression for z into ${ \mathbf{F} }$, we get:

${ \mathbf{F}(x, y) = (x+y)\mathbf{i} + (2x - (6 - 3x - 2y))\mathbf{j} + (y + (6 - 3x - 2y))\mathbf{k} }$

Simplifying the components, we have:

${ \mathbf{F}(x, y) = (x+y)\mathbf{i} + (5x + 2y - 6)\mathbf{j} + (-3x - y + 6)\mathbf{k} }$

This expression gives the vector field ${ \mathbf{F} }$ at any point on the surface S in terms of the parameters x and y. Now, we can compute the dot product of ${ \mathbf{F} }$ and the normal vector ${ \mathbf{n} }$, which is a crucial step in evaluating the surface integral. By expressing ${ \mathbf{F} }$ in terms of x and y, we make it compatible with the parameterization of the surface, allowing us to integrate over the region D in the xy-plane.

Step 4: Computing the Dot Product ${ \mathbf{F} \cdot \mathbf{n} }$

Now that we have both the vector field ${ \mathbf{F}(x, y) = (x+y)\mathbf{i} + (5x + 2y - 6)\mathbf{j} + (-3x - y + 6)\mathbf{k} }$ and the normal vector ${ \mathbf{n} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} }$, we can compute their dot product. The dot product ${ \mathbf{F} \cdot \mathbf{n} }$ is given by:

${ \mathbf{F} \cdot \mathbf{n} = (x+y)(3) + (5x + 2y - 6)(2) + (-3x - y + 6)(1) }$

Expanding and simplifying this expression, we get:

${ \mathbf{F} \cdot \mathbf{n} = 3x + 3y + 10x + 4y - 12 - 3x - y + 6 }$

${ \mathbf{F} \cdot \mathbf{n} = 10x + 6y - 6 }$

This dot product represents the component of the vector field ${ \mathbf{F} }$ that is normal to the surface S. It is a scalar function in terms of x and y, and it is the integrand in our surface integral. With ${ \mathbf{F} \cdot \mathbf{n} }$ computed, we are now ready to set up and evaluate the double integral over the region D in the xy-plane. The dot product simplifies the process by combining the vector nature of ${ \mathbf{F} }$ and the surface orientation into a single scalar function.

Step 5: Setting Up and Evaluating the Double Integral

We are now prepared to evaluate the surface integral. The surface integral ${ \iint_S \mathbf{F} \cdot d\mathbf{s} }$ can be expressed as a double integral over the region D in the xy-plane:

${ \iint_S \mathbf{F} \cdot d\mathbf{s} = \iint_D (\mathbf{F} \cdot \mathbf{n}) dA }$

where ${ dA = dx dy }$ (or ${ dy dx }$ depending on the order of integration). We found that ${ \mathbf{F} \cdot \mathbf{n} = 10x + 6y - 6 }$, and the region D is defined by ${ 0 \leq x \leq 2 }$ and ${ 0 \leq y \leq 3 - \frac{3}{2}x }$. Therefore, the double integral is:

${ \iint_D (10x + 6y - 6) dA = \int_{0}^{2} \int_{0}^{3 - \frac{3}{2}x} (10x + 6y - 6) dy dx }$

First, we integrate with respect to y:

${ \int_{0}^{3 - \frac{3}{2}x} (10x + 6y - 6) dy = \left[ 10xy + 3y^2 - 6y \right]_{0}^{3 - \frac{3}{2}x} }$

Plugging in the limits of integration for y, we get:

${ 10x\left(3 - \frac{3}{2}x\right) + 3\left(3 - \frac{3}{2}x\right)^2 - 6\left(3 - \frac{3}{2}x\right) }$

Expanding and simplifying this expression:

${ 30x - 15x^2 + 3\left(9 - 9x + \frac{9}{4}x^2\right) - 18 + 9x }$

${ 30x - 15x^2 + 27 - 27x + \frac{27}{4}x^2 - 18 + 9x }$

${ -\frac{33}{4}x^2 + 12x + 9 }$

Now, we integrate with respect to x:

${ \int_{0}^{2} \left( -\frac{33}{4}x^2 + 12x + 9 \right) dx = \left[ -\frac{11}{4}x^3 + 6x^2 + 9x \right]_{0}^{2} }$

Plugging in the limits of integration for x, we get:

${ -\frac{11}{4}(2)^3 + 6(2)^2 + 9(2) = -\frac{11}{4}(8) + 6(4) + 18 }$

${ -22 + 24 + 18 = 20 }$

Thus, the surface integral ${ \iint_S \mathbf{F} \cdot d\mathbf{s} = 20 }$.

Conclusion

We have successfully evaluated the surface integral ${ \iint_S \mathbf{F} \cdot d\mathbf{s} }$ for the given vector field ${ \mathbf{F} }$ and surface S. The process involved parameterizing the surface, computing the normal vector, expressing the vector field in terms of the parameters, computing the dot product ${ \mathbf{F} \cdot \mathbf{n} }$, and finally, evaluating the double integral. The result, 20, represents the flux of the vector field ${ \mathbf{F} }$ across the surface S. This comprehensive step-by-step approach provides a clear methodology for solving similar problems in vector calculus, highlighting the importance of each step in achieving the final result. Understanding these techniques is invaluable for applications in physics, engineering, and other fields where vector fields and surface integrals are prevalent.