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. \(\mathbf{F}=\mathbf{r} /|\mathbf{r}|^{3}\) across the sphere of radius \(a\) centered at the origin, where \(\mathbf{r}=\langle x, y, z\rangle\); normal vectors point outward.

Short answer

Based on the step-by-step solution provided, answer the following question: **Question**: Calculate the flux of the vector field \(\mathbf{F} = \frac{\mathbf{r}}{|\mathbf{r}|^{3}}\) across the sphere of radius \(a\). **Answer**: The flux of the vector field across the sphere of radius \(a\) is \(\frac{8}{3}a^2\pi\).

Step-by-step solution

Parametrization of the sphere

Given a sphere of radius \(a\) centered at the origin, its parametric equation can be written as: $$\mathbf{r}(u,v) = \langle a\sin{u}\cos{v}, a\sin{u}\sin{v}, a\cos{u}\rangle$$ for \(u\in[0,\pi]\) and \(v\in[0,2\pi]\).

Calculate the tangent vectors

To compute the surface element \(dS\), we need the tangent vectors with respect to \(u\) and \(v\): $$\frac{\partial \mathbf{r}}{\partial u}=\langle a\cos{u}\cos{v}, a\cos{u}\sin{v}, -a\sin{u}\rangle$$ and $$\frac{\partial \mathbf{r}}{\partial v}= \langle -a\sin{u}\sin{v}, a\sin{u}\cos{v}, 0\rangle$$

Find the normal vector

To get the outward-pointing normal vector, we take the cross product of the tangent vectors: $$\mathbf{n}=\frac{\partial \mathbf{r}}{\partial u}\times\frac{\partial \mathbf{r}}{\partial v} = \langle a^2\sin^2{u}\cos{u}, a^2\sin^2{u}\sin{u}, a^2\sin{u}\cos{u}\rangle$$

Determine the given vector field in terms of the parametrization

We are given the vector field: $$\mathbf{F}=\frac{\mathbf{r}}{|\mathbf{r}|^{3}}$$ In terms of the parametric representation of \(\mathbf{r}\), we can rewrite \(\mathbf{F}\) as: $$\mathbf{F}(u,v)=\frac{1}{a^2} \langle a\sin{u}\cos{v}, a\sin{u}\sin{v}, a\cos{u}\rangle$$

Compute the flux integral

Now we are ready to compute the flux integral: $$\iint\limits_S \mathbf{F} \cdot \mathbf{n} dS = \iint\limits_D \mathbf{F}(u,v) \cdot (\frac{\partial \mathbf{r}}{\partial u}\times\frac{\partial \mathbf{r}}{\partial v}) dudv$$ $$= \int_{0}^{\pi} \int_{0}^{2\pi} \left(\frac{1}{a^2} \langle a\sin{u}\cos{v}, a\sin{u}\sin{v}, a\cos{u}\rangle \cdot \langle a^2\sin^2{u}\cos{u}, a^2\sin^2{u}\sin{u}, a^2\sin{u}\cos{u}\rangle\right) dudv$$ $$= \int_{0}^{\pi} \int_{0}^{2\pi} a^2\sin^3{u}\cos^2{v}+a^2\sin^3{u}\sin^2{v}+a^2\sin{u}\cos^2{u} dudv$$ This simplifies to: $$= a^2 \int_{0}^{\pi} \int_{0}^{2\pi} \sin^3{u} dudv = a^2 \left(\int_{0}^{\pi} \sin^3{u} du\right) \left(\int_{0}^{2\pi} dv\right)$$ We can solve the integral as follows: $$ \int_{0}^{\pi} \sin^3{u} du = \frac{4}{3}$$ $$ \int_{0}^{2\pi} dv = 2\pi$$ Therefore, the final result is: $$\text{Flux} = a^2 \times \frac{4}{3} \times 2\pi = \boxed{\frac{8}{3}a^2\pi}$$

Key concepts

Flux

Flux measures the quantity of a vector field that passes through a surface. Imagine a net placed in a stream, capturing water flowing through it. The amount of water caught represents the flux.

In mathematical terms, for a vector field \( \mathbf{F} \), the flux through a surface \( S \) is expressed by the surface integral:

• \( \iint \limits_S \mathbf{F} \cdot d\mathbf{S} \)

Here, \( d\mathbf{S} \) is an oriented surface element with a direction given by the surface normal vector.

Flux can be seen as the dot product of the vector field and the normal vector over each point on the surface. A positive flux indicates that the field direction and the surface orientation agree, whereas a negative flux shows they oppose each other.

Vector Fields

A vector field assigns a vector to every point in space. It's like having a direction and magnitude at each point in the area you're looking at.

To understand this, think of a weather map where each point shows wind's speed and direction. In the exercise, the vector field is given by:

• \( \mathbf{F} = \frac{\mathbf{r}}{\|\mathbf{r}\|^3} \)

This expression represents a radial field emanating from the origin.

For the sphere problem, the vector field is re-expressed in terms of the parametric variables. This helps in simplifying calculations since these variables better describe the sphere's geometry.

Parametric Equations

Parametric equations describe surfaces using parameters, simplifying the representation of complex shapes.

For a sphere centered at the origin with radius \( a \), the parametric equations provided are:

• \( \mathbf{r}(u,v) = \langle a\sin{u}\cos{v}, a\sin{u}\sin{v}, a\cos{u} \rangle \)

Here, \( u \) and \( v \) vary within specific intervals to cover the entire sphere.

Such representations are crucial. They enable the transformation of surface integrals into simpler double integrals over a parameter plane. By expressing surfaces parametrically, complex spatial problems become manageable through standard calculus techniques.

Normal Vector

The normal vector is perpendicular to a surface. It points in the direction "straight out" from the surface at any point.

Finding the normal vector involves calculating the cross product of two tangent vectors. For the sphere, the tangent vectors with respect to \( u \) and \( v \) are derived first, and their cross product yields the outward normal vector:

• \( \mathbf{n} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \)

This vector's orientation aligns with how the surface is defined: outward in this case.

The normal vector plays a pivotal role in computing surface integrals, as it determines the direction of the surface orientation. The flux calculation explicitly uses the normal vector to evaluate how much of the vector field actually 'crosses' the surface.