Question

Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an explicit or parametric description of the surface. \(\mathbf{F}=\left\langle e^{-y}, 2 z, x y\right\rangle\) across the curved sides of the surface \(S=\\{(x, y, z): z=\cos y,|y| \leq \pi, 0 \leq x \leq 4\\} ;\) normal vectors point upward.

Short answer

To find the flux of the vector field \(\mathbf{F}=\left\langle e^{-y}, 2z, xy\right\rangle\) across the given surface \(S=\{(x, y, z): z=\cos y, |y| \leq \pi, 0 \leq x \leq 4\}\) and with upward-pointing normal vectors, we first represent the surface using the position vector \(\mathbf{r}(x, y) = \left\langle x, y, \cos y \right\rangle\). Then, we find the surface normal \(\mathbf{n}\) by taking the cross product of the partial derivatives with respect to \(x\) and \(y\), obtaining \(\mathbf{n}=\left\langle 0, \sin y, 1 \right\rangle\). Next, we compute the dot product of the vector field and the surface normal \(\mathbf{F}\cdot \mathbf{n} = 2\cos y\sin y + xy\). Finally, the surface integral over the domain of the parameters \(0\le x \le 4\) and \(-\pi\le y\le \pi\) gives the flux as 0.

Step-by-step solution

Find the position vector \(\mathbf{r}(x, y)\)

Since the surface \(S\) is given by \(z = \cos y\) and \(|y| \leq \pi\), \(0 \leq x \leq 4\); we can represent the surface using the position vector \(\mathbf{r}(x, y) = \left\langle x, y, \cos y \right\rangle\).

Calculate the surface normal \(\mathbf{n}\)

Compute the partial derivatives of the position vector with respect to \(x\) and \(y\). \(\frac{\partial \mathbf{r}}{\partial x} = \left\langle 1, 0, 0 \right\rangle\) \(\frac{\partial \mathbf{r}}{\partial y} = \left\langle 0, 1, -\sin y \right\rangle\) Next, calculate the cross product of these partial derivatives to find the surface normal \(\mathbf{n}\): \(\mathbf{n} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} = \left\langle 1, 0, 0 \right\rangle \times \left\langle 0, 1, -\sin y \right\rangle = \left\langle 0, \sin y, 1 \right\rangle\)

Compute \(\mathbf{F}\cdot \mathbf{n}\)

Substitute \(\mathbf{r}(x, y)\) into the given vector field \(\mathbf{F}\). \(\mathbf{F}(\mathbf{r}(x,y))=\left\langle e^{-y}, 2\cos y, xy\right\rangle\) Now, find the dot product of \(\mathbf{F}\) and \(\mathbf{n}\): \(\mathbf{F}\cdot \mathbf{n} = \left\langle e^{-y}, 2\cos y, xy\right\rangle \cdot \left\langle 0, \sin y, 1 \right\rangle = 2\cos y\sin y + xy\)

Compute the surface integral

To find the flux of the vector field across the surface, we need to compute the surface integral: \(\iint_S \mathbf{F}\cdot d\mathbf{S}=\iint_{D} \mathbf{F}\cdot \mathbf{n} \, dx\, dy\) where \(D\) is the domain of the parameters \(x\) and \(y\). From the problem description, the domain of the parameters is \(0\le x \le 4\) and \(-\pi\le y\le \pi\), which is a rectangle. So the integral is: \(\iint_{D} (2\cos y\sin y + xy) \, dx\, dy= \int_{-\pi}^{\pi} \int_{0}^{4} (2\cos y\sin y + xy) \, dx\, dy\) Let's first evaluate the inner integral with respect to \(x\): \(\int_{0}^{4} (2\cos y\sin y + xy) \, dx = \left[2\cos y\sin yx + \frac{1}{2}x^2y\right]_{0}^{4}\) = \(8\cos y\sin y + 8y\) Now, evaluate the outer integral with respect to \(y\): \(\int_{-\pi}^{\pi} (8\cos y\sin y +8y) \, dy\) Since \(\cos y \sin y\) and \(y\) are odd functions, their integrals over a symmetric interval from \(-\pi\) to \(\pi\) will be zero: \(\int_{-\pi}^{\pi} (8\cos y\sin y + 8y) \, dy = 0\) Therefore, the flux of the vector field across the surface is 0.

Key concepts

Surface Parametrization

When dealing with flux calculations in vector fields, surface parametrization plays a crucial role. It allows us to transform a complex surface into manageable mathematics that we can work with. Consider a surface defined by the set of equations and conditions as given in the problem: the curved surface where \(z = \cos y\), with constraints \(|y| \leq \pi\) and \(0 \leq x \leq 4\). To represent this surface, we use a position vector \(\mathbf{r}(x, y) = \langle x, y, \cos y \rangle\). This vector provides a mapping from the \((x, y)\) parameter space to the 3D surface.

Using these parameters, the complex surface is now translated into a simpler 2D representation within the parameter space. This simplification enables us to compute further mathematical operations necessary for flux calculations, such as cross products and integrals. Essentially, parametrization turns geometric complexities into algebraic expressions.

Cross Product

In vector calculus, the cross product is a vital tool for finding the direction that is perpendicular to two vectors. For our surface, this involves calculating a surface normal vector, which is necessary for determining flux. To compute the normal, we first need partial derivatives of our position vector \(\mathbf{r}(x, y)\) with respect to the parameters \(x\) and \(y\).

• \(\frac{\partial \mathbf{r}}{\partial x} = \langle 1, 0, 0 \rangle\)
• \(\frac{\partial \mathbf{r}}{\partial y} = \langle 0, 1, -\sin y \rangle\)

The cross product of these vectors \(\frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y}\) gives us the normal vector:\
\(\mathbf{n} = \langle 0, \sin y, 1 \rangle\).

This vector is crucial for calculating flux because it represents the orientation of the surface at any point \((x, y)\), and will be used in the surface integral as part of the dot product operation.

Surface Integral

A surface integral allows you to calculate the total flux passing through a surface within a vector field. Flux quantifies the amount of a vector field passing through a given surface. We use surface integrals to find this quantity. Leveraging the parametric description of a surface, we can compute:

\[\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_{D} \mathbf{F} \cdot \mathbf{n} \, dx\, dy\]

where \(D\) is the domain defined by \(0 \leq x \leq 4\) and \(-\pi \leq y \leq \pi\). This integral describes the total flux across the surface \(S\). The product \(\mathbf{F} \cdot \mathbf{n}\) gives us a scalar field representing the flow through a unit of area perpendicular to the surface at each point in the domain.

• First, compute the inner integral with respect to \(x\)
• Next, evaluate the outer integral with respect to \(y\)

After these operations, you end up with a clear answer for the total flux, as demonstrated in the step-by-step solution.

Dot Product

The dot product operation is central to flux calculations within vector fields. It is used to measure how much of the vector field penetrates through the surface at each point. For our problem, we calculate the dot product between the vector field \(\mathbf{F}\) evaluated at \(\mathbf{r}(x,y)\) and the normal vector \(\mathbf{n}\). This gives the expression:

\(\mathbf{F} \cdot \mathbf{n} = 2\cos y\sin y + xy\).

The dot product gives us a scalar that quantifies the degree to which the field \(\mathbf{F}\) aligns with the surface normal \(\mathbf{n}\) at every point. If the dot product is positive, the field is leaving the surface; if negative, it's entering - and a zero value indicates tangential flow. This scalar function is then used in the surface integral to calculate the total flux across the surface. Thus, the dot product serves as a bridge between the vector nature of \(\mathbf{F}\) and the scalar nature of flux.