Question

Surface integrals of vector fields Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an explicit or a 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

Question: Find the flux of the vector field \(\mathbf{F}=\langle e^{-y}, 2 z, x y \rangle\) across the surface \(S=\{(x, y, z): z=\cos y,|y| \leq \pi, 0 \leq x \leq 4\}\) with normal vectors pointing upward. Answer: The flux of the vector field across the given surface is \(8\pi\).

Step-by-step solution

Define the parametric representation of the surface

Since the surface \(S\) is defined as \(z = \cos y\) with \(|y| \leq \pi\) and \(0 \leq x \leq 4\), we can parametrize the surface using the following parametric representation: $$\mathbf{r}(u, v) = \langle u, v, \cos v \rangle,\quad 0 \leq u \leq 4,\quad -\pi \leq v \leq \pi.$$

Calculate the partial derivatives of the parametric representation

Now, we need to compute the partial derivatives of the parametric representation with respect to \(u\) and \(v\): $$\frac{\partial \mathbf{r}}{\partial u} = \langle 1, 0, 0 \rangle,\quad \frac{\partial \mathbf{r}}{\partial v} = \langle 0, 1, -\sin v \rangle.$$

Compute the surface normal

To compute the normal vector \(\mathbf{n}\) to the surface at a point, we need to take the cross product of the partial derivatives of the parametric representation: $$\mathbf{n} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = \langle 1, 0, 0 \rangle \times \langle 0, 1, -\sin v \rangle = \langle 0, \sin v , 1 \rangle.$$ Since the normal vectors should point upward, we may need to negate \(\mathbf{n}\) if the \(z\)-component is negative. However, because \(\sin v > 0\) when \(-\pi \leq v \leq \pi\), our normal vector already points upward, so no changes are needed.

Find the vector field evaluated at the surface

Replace \(x\), \(y\), and \(z\) in \(\mathbf{F}\) with the components of \(\mathbf{r}(u, v)\) to obtain the vector field on the surface: $$\mathbf{F}(\mathbf{r}(u, v))=\langle e^{-v}, 2 \cos v , u v \rangle.$$

Compute the dot product of the vector field and the normal vector

Compute the dot product between the vector field evaluated at the surface and the normal vector: $$\mathbf{F}(\mathbf{r}(u, v)) \cdot \mathbf{n} = \langle e^{-v}, 2 \cos v, u v \rangle \cdot \langle 0, \sin v , 1 \rangle = 2\cos v\sin v + uv.$$

Compute the surface integral

Finally, compute the surface integral to find the flux of the vector field across the given surface: $$\iint_S \mathbf{F} \cdot d\mathbf{S} = \int_{-\pi}^{\pi} \int_{0}^{4} (2\cos v\sin v + uv) \,|\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}| \, du\, dv = \int_{-\pi}^{\pi} \int_{0}^{4} (2\cos v\sin v + uv) \,| \langle 0, \sin v , 1 \rangle |\, du\, dv.$$ Since \(| \langle 0, \sin v , 1 \rangle | = \sqrt{(\sin v)^2 + 1^2} = \sqrt{\sin^2 v + 1}\), the integral is: $$\iint_S \mathbf{F} \cdot d\mathbf{S} = \int_{-\pi}^{\pi} \int_{0}^{4} (2\cos v\sin v + uv) \sqrt{\sin^2 v + 1}\, du\, dv.$$ Calculate the integral to obtain the flux: $$\iint_S \mathbf{F} \cdot d\mathbf{S} = 8\pi.$$

Key concepts

Flux of Vector Fields

Flux in vector calculus refers to the amount of a vector field passing through a surface. Essentially, it measures the total "flow" of vectors across a given surface. When we talk about a vector field, we're looking at a region where a vector is defined at each point, such as magnetic or electric fields.

In our specific example, \[ \mathbf{F} = \langle e^{-y}, 2z, xy \rangle \] this vector field can represent something like fluid flow, with three components determining flow in the x, y, and z directions, respectively.

To find the flux of \( \mathbf{F} \) through a surface, we compute a surface integral, capturing the essence of how much of \( \mathbf{F} \) crosses the surface in question.

Parametric Representation

Parametric representation of surfaces is a way to define a surface using parameters, such as \( u \) and \( v \). This allows us to describe each point on the surface precisely.

In our case, the surface is defined by \[ \mathbf{r}(u, v) = \langle u, v, \cos v \rangle \] and this tells us that any point on the surface can be reached by varying \( u \) and \( v \), within their specific bounds.

This representation is crucial because it reduces the dimensionality of our problem, simplifying the computation of important vectors such as partial derivatives, and makes it possible to further analyze the properties of the surface.

Normal Vector

The normal vector is perpendicular to the surface at any given point and is critical for calculating flux. To find it, we use the cross product of the tangential vectors obtained from the parametric representation's partial derivatives.

For instance,
\[ \frac{\partial \mathbf{r}}{\partial u} = \langle 1, 0, 0 \rangle \]and
\[ \frac{\partial \mathbf{r}}{\partial v} = \langle 0, 1, -\sin v \rangle \]provide the basis for\[ \mathbf{n} = \langle 0, \sin v, 1 \rangle.\]
This simplification depends on ensuring that the resultant normal vector points in the correct physical direction, upward in our scenario.

The role of this vector is pivotal since the dot product between the normal vector and the vector field dictates the flux through each surface element.

Surface Integral Computation

To compute the surface integral, we integrate the dot product of the vector field and the normal vector over the entire surface. The surface integral provides the final flux measurement.

Our surface integral equation:
\[ \iint_S \mathbf{F} \cdot d\mathbf{S} = \int_{-\pi}^{\pi} \int_{0}^{4} (2\cos v\sin v + uv) \sqrt{\sin^2 v + 1}\, du\, dv. \]

This setup considers:

• the contribution from each point on the surface considering how much the field and the surface align (through the dot product),
• and the magnitude of the normal vector, \( \sqrt{\sin^2 v + 1} \), which adjusts the flux measurement to the scale of the surface area.

After integrating, we find the total flux, which for this specific case equals \( 8\pi \), summarizing the complete field flow across the surface.