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A scalar-valued function \(\varphi\) is harmonic on a region \(D\) if \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=0\) at all points of \(D\). Show that if \(\varphi\) is harmonic on a region \(D\) enclosed by a surface \(S\) then \(\iint_{S} \nabla \varphi \cdot \mathbf{n} d S=0\)

Short Answer

Expert verified
Question: Prove that if a scalar-valued function φ is harmonic on a region D, then the surface integral of its gradient dotted with the normal vector over the enclosing surface S is zero. Answer: By applying the Divergence Theorem and given that the Laplacian of the scalar function φ is zero (i.e., ∇²φ = 0), we have shown that the surface integral of the gradient of φ dotted with the normal vector over the surface S is indeed equal to zero. Mathematically, it is represented as ∮∮ₛ ∇φ ⋅ 𝐧 dS = 0.

Step by step solution

01

Recall the Divergence Theorem

The Divergence Theorem tells us that for a vector field, \(\textbf{F}\), defined over a region \(D\) enclosed by a surface \(S\), the volume integral of the divergence of \(\textbf{F}\) is equal to the surface integral of the field over the surface. Mathematically, it can be written as: $$\iiint_D (\nabla \cdot \textbf{F})dV = \iint_{S} \textbf{F} \cdot \mathbf{n} dS$$ where \(\mathbf{n}\) is the unit normal vector on the surface \(S\), pointing outward, and \(dV\) and \(dS\) are the infinitesimal volume and surface elements, respectively.
02

Apply the Divergence Theorem to the exercise

Let's apply the divergence theorem to the gradient of \(\varphi\). We are given that \(\nabla^2 \varphi = \nabla \cdot (\nabla \varphi) = 0\). So we have: $$\iiint_D (\nabla \cdot (\nabla \varphi))dV = \iint_{S} (\nabla \varphi) \cdot \mathbf{n} dS$$ Since \(\nabla^2 \varphi = 0\), the left side of the equation becomes: $$\iiint_D 0 \, dV = 0$$ So, we get: $$0 = \iint_{S} (\nabla \varphi) \cdot \mathbf{n} dS$$
03

Conclusion

We have shown that for a harmonic function \(\varphi\), with \(\nabla^2 \varphi=0\) in a region \(D\) enclosed by a surface \(S\), the surface integral of its gradient dotted with the unit normal vector is zero: $$\iint_{S} \nabla \varphi \cdot \mathbf{n} d S=0$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divergence Theorem
The Divergence Theorem is a key concept in vector calculus. It connects the flow (or divergence) of a vector field through a three-dimensional region to the flow across its boundary. This theorem is useful for converting volume integrals into surface integrals.
  • Mathematically, the theorem states: \[ \iiint_D (abla \cdot \textbf{F}) \,dV = \iint_{S} \textbf{F} \cdot \mathbf{n} \,dS \]
  • Here, \(\textbf{F}\) is a vector field defined in region \(D\), \(\mathbf{n}\) is the outward unit normal on boundary \(S\), and \(dV\) and \(dS\) are volume and surface elements.
The Divergence Theorem links the total "outflow" of a vector component directly from a region to the sum of the vector field over the region's boundary. This provides a powerful tool in physics and engineering, particularly in fluid dynamics and electromagnetism, to simplify calculations.
Scalar-Valued Functions
Scalar-valued functions assign a single value to every point in a space. Unlike vector-valued functions, which yield a vector, scalar-valued functions result in a singular dimension numerical output.
  • An example of a scalar-valued function is the temperature distribution in a room, where each location correlates with a specific temperature value.
  • These functions can be complex and multidimensional, depending solely on the number of variables involved.
  • In the context of harmonic functions, which are a subset of scalar-valued functions, their Laplacian \(abla^2 \varphi\) equals zero. This means they lack any "source" or "sink" within their domain, making them particularly useful in fields like electrostatics or potential flow.
Understanding scalar-valued functions is important because they simplify the analysis of phenomena that are inherently scalar, like potential fields or temperature.
Surface Integrals
Surface integrals are crucial in calculus for analyzing the flow across surfaces or through regions. They allow us to calculate how a vector field interacts with a surface.
  • The basic formulation of a surface integral is \(\iint_{S} \textbf{F} \cdot \mathbf{n} \,dS\), where \(\textbf{F}\) is the vector field and \(S\) is the surface over which the integration is conducted.
  • \(\mathbf{n}\) is an important component here as it represents the unit normal to the surface at each point, guiding the directionality of the flow or flux.
  • Surface integrals are indispensable in many practical applications such as calculating the flux of magnetic fields over a surface or the rate of fluid flow through a pipe's surface.
By working with surface integrals, students and professionals can better understand and measure the interaction between fields and surfaces, giving insights into various physical processes.

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Most popular questions from this chapter

Let \(\mathbf{F}=\langle z, 0,-y\rangle\) a. Find the scalar component of curl \(\mathbf{F}\) in the direction of the unit vector \(\mathbf{n}=\langle 1,0,0\rangle\). b. Find the scalar component of curl \(\mathbf{F}\) in the direction of the unit vector \(\mathbf{n}=\left\langle\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle\). c. Find the unit vector \(\mathbf{n}\) that maximizes \(\operatorname{scal}_{\mathbf{n}}\langle-1,1,0\rangle\) and state the value of \(\operatorname{scal}_{\mathbf{n}}\langle-1,1,0\rangle\) in this direction.

Average circulation Let \(S\) be a small circular disk of radius \(R\) centered at the point \(P\) with a unit normal vector \(\mathbf{n}\). Let \(C\) be the boundary of \(S\). a. Express the average circulation of the vector field \(\mathbf{F}\) on \(S\) as a surface integral of \(\nabla \times \mathbf{F}\) b. Argue that for small \(R\), the average circulation approaches \(\left.(\nabla \times \mathbf{F})\right|_{P} \cdot \mathbf{n}(\text { the component of } \nabla \times \mathbf{F} \text { in the direction of } \mathbf{n}\) evaluated at \(P\) ) with the approximation improving as \(R \rightarrow 0\)

Find the general formula for the surface area of a cone with height \(h\) and base radius \(a\) (excluding the base).

Fourier's Law of heat transfer (or heat conduction ) states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows from hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of \(J /(m-s-K)\) A temperature function for a region \(D\) is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of \(D\) In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume \(k=1 .\) \(T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}} ; D\) is the sphere of radius \(a\) centered at the origin.

One of Maxwell's equations for electromagnetic waves is \(\nabla \times \mathbf{B}=C \frac{\partial \mathbf{E}}{\partial t},\) where \(\mathbf{E}\) is the electric field, \(\mathbf{B}\) is the magnetic field, and \(C\) is a constant. a. Show that the fields \(\mathbf{E}(z, t)=A \sin (k z-\omega t) \mathbf{i}\) and \(\mathbf{B}(z, t)=A \sin (k z-\omega t) \mathbf{j}\) satisfy the equation for constants \(A, k,\) and \(\omega,\) provided \(\omega=k / C\). b. Make a rough sketch showing the directions of \(\mathbf{E}\) and \(\mathbf{B}\).

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