Chapter 9: Problem 5
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 \((1,0,0),(0,1,0),(0,0,1)\)
Short Answer
Step by step solution
Understanding Stokes' Theorem
Determine the Curl of \(\mathbf{F}\)
Parameterize the Surface
Determine the Surface Element \(d\mathbf{S}\)
Evaluate the Surface Integral
Solve the Double Integral
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
curl of a vector field
To compute the curl of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), we use the formula:
- \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \).
line integral
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
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
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 \( x + y + z = 1 \).
By parameterizing the surface effectively, complex geometric boundaries become simpler, aiding in the calculation of integrals and deeper understanding of the geometry involved.