Problem 5 - Advanced Engineering Mathematics

Question

In Problems \(5-12\), use Stokes' theorem to evaluate \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\). Assume \(C\) is oriented counterclockwise as viewed from above. \(\mathbf{F}=(2 z+x) \mathbf{i}+(y-z) \mathbf{j}+(x+y) \mathbf{k} ; C\) the triangle with vertices \((1,0,0),(0,1,0),(0,0,1)$$\mathbf{F}=(2 z+x) \mathbf{i}+(y-z) \mathbf{j}+(x+y) \mathbf{k} ; C\) the triangle with vertices \((1,0,0),(0,1,0),(0,0,1)\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{3}{2}\).

Step 1: Understanding Stokes' Theorem

Stokes' Theorem relates a line integral around a closed curve \(C\) to a surface integral over a surface \(S\) that is bounded by \(C\). The theorem is given by \(\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S}\), where \(abla \times \mathbf{F}\) is the curl of \(\mathbf{F}\).

Step 2: Determine the Curl of \(\mathbf{F}\)

Calculate the curl of \(\mathbf{F}=(2z+x)\mathbf{i}+(y-z)\mathbf{j}+(x+y)\mathbf{k}\). The curl is given by:\[abla \times \mathbf{F} = \left(\frac{\partial (x+y)}{\partial y} - \frac{\partial (y-z)}{\partial z}\right) \mathbf{i} - \left(\frac{\partial (x+y)}{\partial x} - \frac{\partial (2z+x)}{\partial z}\right) \mathbf{j} + \left(\frac{\partial (y-z)}{\partial x} - \frac{\partial (2z+x)}{\partial y}\right) \mathbf{k}\].Substituting the partial derivatives, we get \(abla \times \mathbf{F} = \mathbf{i} + \mathbf{j} + \mathbf{k}\).

Step 3: Parameterize the Surface

The surface \(S\) is a triangle with vertices \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\). We can parameterize \(S\) using parameters \(u\) and \(v\) such that \((x,y,z) = (1-u-v,u,v)\); thus, the surface equation is \(x+y+z = 1\).

Step 4: Determine the Surface Element \(d\mathbf{S}\)

Since \(S\) is a plane, \(d\mathbf{S} = \mathbf{n}\,dS\) where \(\mathbf{n}\) is a unit normal vector to the surface. For the given plane, \(\mathbf{n} = \mathbf{i} + \mathbf{j} + \mathbf{k}\), and the area element is \(dS = dudv\). Hence, \(d\mathbf{S} = (\mathbf{i} + \mathbf{j} + \mathbf{k})dudv\).

Step 5: Evaluate the Surface Integral

Substitute \(abla \times \mathbf{F} = \mathbf{i} + \mathbf{j} + \mathbf{k}\) and \(d\mathbf{S} = (\mathbf{i} + \mathbf{j} + \mathbf{k})dudv\) in the surface integral:\[\iint_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_{D} (1+1+1) dudv = \iint_{D} 3 dudv,\]where \(D\) is the triangular region in the \(uv\) plane where \(0 \leq u \leq 1\) and \(0 \leq v \leq 1-u\).

Step 6: Solve the Double Integral

Evaluate the integral:\[\iint_{D} 3 dudv = \int_{0}^{1} \int_{0}^{1-u} 3 \, dv \, du = \int_{0}^{1} [3v]_{0}^{1-u} du = \int_{0}^{1} 3(1-u) \, du.\]Solve the integral:\[= 3\left[u - \frac{u^2}{2}\right]_{0}^{1} = 3\left(1-\frac{1}{2}\right) = 3 \times \frac{1}{2} = \frac{3}{2}.\]

Step 7: Conclusion

The value of the line integral \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\) using Stokes' Theorem is \(\frac{3}{2}\).

Key Concepts

curl of a vector field

To understand Stokes' Theorem, we need to first grasp the concept of the "curl of a vector field". It is a vector operation describing the infinitesimal rotation of a 3D vector field. Think of it like this: if a vector field represents the velocity of a fluid flow, the curl measures how much and in what direction the fluid is rotating at a point.

To compute the curl of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), we use the formula:

• \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \).

For the given function \( \mathbf{F} = (2z+x) \mathbf{i} + (y-z) \mathbf{j} + (x+y) \mathbf{k} \), calculating these partial derivatives and substituting them, we find that \( abla \times \mathbf{F} = \mathbf{i} + \mathbf{j} + \mathbf{k} \). This provides a direction and magnitude of rotation for our vector field.

line integral

The concept of a "line integral" is important when dealing with vector fields. It measures how a vector field \( \mathbf{F} \) influences a path \( C \), providing insight into the work done by the field along that path.

In mathematical terms, the line integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \) can be visualized by breaking down the path into infinitesimal segments. Each segment has a vector \( d\mathbf{r} \) tangential to the path. The work done by \( \mathbf{F} \) along each segment is \( \mathbf{F} \cdot d\mathbf{r} \), meaning the dot product of the field and the path. When all these contributions along the curve \( C \) are summed, we obtain the total line integral.

However, directly solving this line integral can be complex. That's where Stokes' Theorem comes in, which connects it to a "surface integral," potentially simplifying computations.

surface integral

A "surface integral" extends the idea of a line integral to higher dimensions. Instead of integrating over a curve, here we integrate over a two-dimensional surface. It effectively measures the flux of a vector field passing through a surface \( S \).

In the context of Stokes' Theorem, a surface integral helps evaluate the flow or rotation (curl) of a vector field across surface \( S \) that \( C \) bounds. For the given problem, where \( abla \times \mathbf{F} = \mathbf{i} + \mathbf{j} + \mathbf{k} \), the surface integral \( \iint_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} \) equals the sum of its dot product with the surface element \( d\mathbf{S} \).

The advantage of utilizing surface integrals in this scenario is that they often simplify to regular double integrals, such as \( \iint_{D} 3 \, dudv \), making them easier to solve while still aligning with the physical interpretation.

parameterization of surfaces

When working with "parameterization of surfaces," it implies transforming the geometrical surface into a form that makes calculus operations, like an integral, more manageable. This is often necessary for evaluating surface integrals.

To parameterize a triangular surface like the one in our problem, involving vertices \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\), we express the position \((x,y,z)\) in terms of new variables, \( u \) and \( v \), that trace out the surface.

• Here, \( (x, y, z) = (1-u-v, u, v) \) captures every point on the plane described by \( x + y + z = 1 \).

Each pair \( (u,v) \) maps to unique \( (x,y,z) \) on the surface, effectively transforming the problem into an integration over \( (u, v) \), which runs through a simpler, clearly defined triangle in the \( uv \)-plane (\( 0 \leq u \leq 1 \) and \( 0 \leq v \leq 1-u \)).
By parameterizing the surface effectively, complex geometric boundaries become simpler, aiding in the calculation of integrals and deeper understanding of the geometry involved.