Chapter 16: Problem 13
Verify that Stokes' Theorem is true for the given vector field \(\mathbf{F}\) and surface \(S\). $$\begin{array}{l}{\mathbf{F}(x, y, z)=y^{2} \mathbf{i}+x \mathbf{j}+z^{2} \mathbf{k}} \\ {S \text { is the part of the paraboloid } z=x^{2}+y^{2} \text { that lies below the }} \\ {\text { plane } z=1, \text { oriented upward }}\end{array}$$
Short Answer
Step by step solution
Stokes' Theorem Overview
Compute \(\nabla \times \mathbf{F}\)
Set Up the Surface Integral
Evaluate Surface Integral
Evaluate the Line Integral
Comparison of Integrals
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Field
- The \( \mathbf{i} \)-component depends on \( y^2 \), meaning it increases with the square of the y-coordinate.
- The \( \mathbf{j} \)-component is simply \( x \), corresponding directly to the x-coordinate.
- The \( \mathbf{k} \)-component is \( z^2 \), increasing with the square of the z-coordinate.
Surface Integral
For our problem, the surface \( S \) is part of a paraboloid, \( z = x^2 + y^2 \), truncated at \( z = 1 \).
- Given surface \( S \) is oriented upward, the surface normal points outward.
- We calculate \( abla \times \mathbf{F} \) to determine how the vector field swirls or rotates over this surface.
- The challenge lies in computing the integral of this curl over \( S \), achieving a relationship with the boundary integral.
Line Integral
For this problem, the boundary curve \( \partial S \) is where the paraboloid cap meets the plane \( z = 1 \).
- We consider this circular boundary located at \( x^2 + y^2 = 1 \).
- Converted into polar coordinates: \( x = \cos(\theta) \), \( y = \sin(\theta) \), maintaining \( z = 1 \).
- The line integral of the vector field \( \mathbf{F} \) explores how \( \mathbf{F} \) aligns with this circular boundary.
Curl of a Vector Field
In the given exercise, the vector field \( \mathbf{F}(x, y, z) = y^{2}\mathbf{i} + x\mathbf{j} + z^{2}\mathbf{k} \) has a curl, \( abla \times \mathbf{F} \), computed as follows:
- The \( \mathbf{i} \)-component: \(0\), as no \( \mathbf{k} \) or \( \mathbf{j} \) terms affect it.
- The \( \mathbf{j} \)-component: \(-1\), representing the field's rotational factor due to \( x \).
- The \( \mathbf{k} \)-component: \(-2y\), representing twisting from the \( y^2 \) term.