Chapter 5: Problem 26
Let \(\mathbf{F}(x, y, z)=x y \mathbf{i}+\left(e^{z^{2}}+y\right) \mathbf{j}+(x+y) \mathbf{k}\) and let \(S\) be the graph of function \(y=\frac{x^{2}}{9}+\frac{z^{2}}{9}-1\) with \(z \leq 0\) oriented so that the normal vector \(S\) has a positive \(y\) component. Use Stokes' theorem to compute integral \(\iint_{S} \operatorname{curl} \mathbf{F} \cdot d \mathbf{S}\).
Short Answer
Step by step solution
Find the curl of F
Verify the surface S
Identify the boundary of S
Apply Stokes' Theorem
Compute the line integral
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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 find the curl of a vector field, we use a determinant involving partial derivatives of the vector field's components. For the vector field \( \mathbf{F}(x, y, z) = x y \mathbf{i} + (e^{z^{2}} + y) \mathbf{j} + (x + y) \mathbf{k} \), the determinant setup looks like:
\[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ x y & e^{z^2} + y & x + y \end{vmatrix} \]
This formula tells us how to calculate the components of the curl vector. After evaluating the determinant, we found that:
\( \operatorname{curl} \mathbf{F} = -2z e^{z^2} \mathbf{i} - \mathbf{j} \).
- The \( \mathbf{i} \) component tells us how much rotation occurs about the x-axis.
- The \( \mathbf{j} \) component tells us about rotation around the y-axis.
- In our example, the \( \mathbf{k} \) component is zero, meaning no rotation around the z-axis.
Surface and Line Integrals
- Surface Integrals: We integrate over a surface, like finding how air flows through a window. For a surface \( S \), and curl vector \( \operatorname{curl} \mathbf{F} \), we find \( \iint_{S} \operatorname{curl} \mathbf{F} \cdot d \mathbf{S} \).
- Line Integrals: These compute the effect along a path or curve, calculated as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \).
Stokes' Theorem establishes an important relationship between these two integrals. It states that the surface integral of the curl of \( \mathbf{F} \) over \( S \) is equal to the line integral of \( \mathbf{F} \) along the boundary \( C \) of that surface:
\[ \iint_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r} \].
This theorem simplifies calculations by transforming surface integrals into line integrals, which are often easier to evaluate. So in the given problem, instead of performing a more complex surface integral, we focus on the curve \( C \) for the line integral.
Parametrization of Curves
For a boundary curve \( C \) that lies in the xy-plane, we require a parametrization to use with line integrals. In the example with \( S \), the boundary occurs where \( z = 0 \), forming a path curve.
Parametrization simplifies the following:
- Representation: We write each point on the curve as \( \mathbf{r}(t) = (x(t), y(t), z(t)) \).
- Domain of \( t \): This is usually an interval like \( [0, 2\pi] \), representing a full circle or ellipse.
For the elliptical path described by our boundary, the parametrization is:
\( \mathbf{r}(t) = (3\cos(t), \cos^2(t) - 1, 0) \),\( 0 \leq t < 2\pi \).
- Advantage: Parametrization turns the complex representation of \( C \) into a simple and manageable form, perfect for substituting into the line integral.
Substituting this \( \mathbf{r}(t) \) into \( \mathbf{F} \), and pairing it with differential elements, we calculate \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), allowing Stokes' theorem to simplify our process!