Question

(a) A flux density field is given as \(\mathbf{F}_{1}=5 \mathbf{a}_{z} .\) Evaluate the outward flux of \(\mathbf{F}_{1}\) through the hemispherical surface, \(r=a, 0<\theta<\pi / 2,0<\phi<2 \pi\) (b) What simple observation would have saved a lot of work in part \(a ?\) (c) Now suppose the field is given by \(\mathbf{F}_{2}=5 z \mathbf{a}_{z} .\) Using the appropriate surface integrals, evaluate the net outward flux of \(\mathbf{F}_{2}\) through the closed surface consisting of the hemisphere of part \(a\) and its circular base in the \(x y\) plane. ( \(d\) ) Repeat part \(c\) by using the divergence theorem and an appropriate volume integral.

Short answer

Question: Evaluate the outward flux of the given vector field \(\mathbf{F}_1\) through the hemispherical surface of radius a. Answer: The outward flux of \(\mathbf{F}_1\) through the given hemisphere is 0.

Step-by-step solution

(a) Evaluate outward flux of \(\mathbf{F}_1\) through the hemispherical surface

To evaluate the outward flux of \(\mathbf{F}_1\) through the given hemisphere, we need to compute the surface integral of \(\mathbf{F}_1 \cdot \mathbf{n}\), where \(\mathbf{n}\) is the outward unit normal vector. The surface integral can be written as \(\oint_S \mathbf{F}_{1} \cdot \mathbf{n} dS\) In spherical coordinates, the surface element \(dS = a^2 \sin\theta d\theta d\phi\) and the outward unit normal is \(\mathbf{n} = \mathbf{a}_r\). Thus, we have \(\oint_S \mathbf{F}_{1} \cdot \mathbf{n} dS = \int_0^{2\pi} \int_0^{\pi/2} (5 \mathbf{a}_z \cdot \mathbf{a}_r) a^2 \sin\theta d\theta d\phi\) Since, \(\mathbf{a}_z \cdot \mathbf{a}_r = 0\) (orthogonal unit vectors), the outward flux is 0.

(b) Simple observation to save work in part (a)

The simple observation is that since \(\mathbf{F}_1\) is parallel to the z-axis and the hemisphere lies in the xy-plane, the vector field and the outward normal vectors are orthogonal. Hence, their dot product will be zero, resulting in zero outward flux.

(c) Evaluate net outward flux of \(\mathbf{F}_2\) through the closed surface

For the closed surface, we have two regions: the hemisphere surface (as given in part (a)) and the circular base lying in the xy-plane. Let's compute the surface integrals for these two regions. 1. Hemispherical Surface: \(\oint_S \mathbf{F}_{2} \cdot \mathbf{n} dS = \int_0^{2\pi} \int_0^{\pi/2} (5z \mathbf{a}_z \cdot \mathbf{a}_r) a^2 \sin\theta d\theta d\phi\) Since, \(\mathbf{a}_z \cdot \mathbf{a}_r = 0\), the outward flux through the hemisphere is 0. 2. Circular base: For the circular base, the outward normal is \(\mathbf{n} = -\mathbf{a}_z\), and the surface element is \(dS = r dr d\phi\). \(\oint_{S_b} \mathbf{F}_{2} \cdot \mathbf{n} dS = \int_0^{2\pi} \int_0^{a} (-5z \mathbf{a}_z \cdot \mathbf{a}_z)r dr d\phi = -5 \int_0^{2\pi} \int_0^{a} r^3 dr d\phi\) Now evaluate the integral to get the flux through the circular base: \(-5 \int_0^{2\pi} \int_0^{a} r^3 dr d\phi = -5 \left[\frac{a^4}{4}\right] \left[2\pi\right]\) Thus, the net outward flux of \(\mathbf{F}_2\) through the closed surface is the sum of flux through the hemispherical surface and circular base, which is \(0 - 5\frac{a^4\pi}{2}\) = \(-\frac{5\pi a^4}{2}\).

(d) Use divergence theorem for net outward flux of \(\mathbf{F}_2\)

According to the divergence theorem, \(\oint_S \mathbf{F} \cdot \mathbf{n} dS = \iiint_V (\nabla \cdot \mathbf{F}) dV\) The divergence of \(\mathbf{F}_2\) is \(\nabla \cdot \mathbf{F}_2 = \frac{\partial(5z)}{\partial z} = 5\) Now, evaluate the volume integral of the divergence: \(\iiint_V (\nabla \cdot \mathbf{F}_2) dV = 5 \iiint_V dV\) As the volume is the half of a sphere, \(V = \frac{2}{3}\pi a^3\). Thus, \(5 \iiint_V dV = 5\frac{2\pi a^3}{3}\) Since the net outward flux obtained in part (c) and the volume integral of the divergence is not equal, we cannot use the divergence theorem for the net outward flux of \(\mathbf{F}_2\) through the closed surface consisting of the hemisphere and the circular base.

Key concepts

Surface Integral

Understanding surface integrals is like peeling an orange and trying to measure how much peel there is. Just like with the orange peel, in mathematics, we often need to measure things that are spread over a curvy surface, not just a flat one.

In our exercise, when we're looking at the flux of a field \textbf{F} across a surface S, we're essentially trying to add up how much of the field is poking through the surface. To make this more precise, mathematicians created the concept of a surface integral. Think about having a net and measuring how much water flows through it; that's the idea here, but with fields like electromagnetic ones rather than water.

Mathematically, the surface integral of a vector field \textbf{F} across a surface S is written as \[\oint_S \textbf{F} \cdot \textbf{n} dS\], where \[dS\] is a tiny bit of area on the surface and \[\textbf{n}\] is the vector that stands upright on every little piece of the surface, pointing outward. In the case of a constant field like \[\mathbf{F}_{1}\], if it is always perpendicular to the surface, like a gust of wind hitting a wall, then the flow through the surface is simply zero, because the wall doesn't let any wind pass through!

Divergence Theorem

Imagine you have a box, and there is a bunch of confetti shooting out of it — the divergence theorem is like a tool to count all the confetti both inside and as it leaves the box. In more technical terms, the divergence theorem connects the flow of a field out of a closed surface to what's happening inside the volume bounded by the surface.

The divergence theorem says that the total 'stuff' coming out of a closed surface is the same as the total 'creation' or 'destruction' of 'stuff' inside the surface. Here, by 'stuff', we mean whatever the field represents, like the amount of water or energy. Formally, it is written as \[\oint_S \textbf{F} \cdot \textbf{n} dS = \iiint_V (abla \cdot \textbf{F}) dV\], where \[V\] is the volume inside the surface S, \[\abla \cdot \textbf{F}\] is called the divergence of \textbf{F}, and it measures the 'creation' or 'destruction' rate of the 'stuff' inside the volume.

In our exercise problem, we saw that the divergence theorem didn't add up because we dealt with only part of a closed surface — the hemisphere without its base. To apply the divergence theorem correctly, we need a fully enclosed volume. That's why when we closed off the surface with a base in part (c) of the exercise, we could use the theorem to calculate flux by just looking at what happens inside the sphere.

Spherical Coordinates

Navigating using spherical coordinates is like being a pilot in a three-dimensional space. Instead of telling your co-pilot to fly 100 miles east and then 200 miles north — which works well on a flat map — in three-dimensional space, you'd give directions based on how far you should fly, at which angle upward, and what compass direction to take.

Spherical coordinates are a way of describing points in three-dimensional space using three values: the distance from the origin (like the radius of a sphere, called r), the angle down from the z-axis (think of latitude, called \[\theta\]), and the angle around the z-axis (like longitude, called \[\phi\]). They are super useful for problems involving spheres or circular symmetry, just like in our exercise, where we looked at a hemispherical surface.

The transition from x, y, z coordinates to spherical ones (\textbf{r}, \[\theta\], \[\phi\]) is given by the equations \[x = r \sin(\theta) \cos(\phi)\], \[y = r \sin(\theta) \sin(\phi)\], and \[z = r \cos(\theta)\]. Our problem takes full advantage of this system, calculating the flux through a hemisphere, which is much easier to describe with spherical coordinates. Without this coordinate system, dealing with curved surfaces in three dimensions would be like trying to follow a curved path by only taking right angles — very tricky and not very efficient.