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Chapter 8: Problem 25

Argue informally that if \(S\) is a closed surface, then $$ \iint_{S}(\nabla \times \mathbf{F}) \cdot d \mathbf{S}=0 $$ (see Exercise 23 ). (A closed surface is one that forms the boundary of a region in space; thus, for example, a sphere is a closed surface.)

Short Answer

Expert verified
For a closed surface, the integral of the curl is zero due to Stokes' Theorem and absence of a boundary.

Step by step solution

01

Understand Stokes' Theorem

Stokes' Theorem relates a surface integral of the curl of a vector field over a surface to a line integral of the vector field over the surface's boundary. Specifically, it states that \(\iint_{S}(abla \times \mathbf{F}) \cdot d \mathbf{S} = \oint_{C} \mathbf{F} \cdot d \mathbf{r}\), where \(C\) is the boundary of the surface \(S\).
02

Identify the Closed Surface Condition

A closed surface, like a sphere or a cube, encloses a volume and does not have a boundary鈥攗nlike an open surface, which has an edge. In the context of Stokes' Theorem, the absence of a boundary \(C\) for a closed surface means that the line integral around the boundary is zero, i.e., \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} = 0\).
03

Apply Stokes' Theorem to the Closed Surface

Given that a closed surface has no boundary, apply Stokes' Theorem to conclude that \(\iint_{S}(abla \times \mathbf{F}) \cdot d \mathbf{S} = \oint_{C} \mathbf{F} \cdot d \mathbf{r} = 0\). This follows directly because the boundary \(C\) does not exist for the closed surface.
04

Conclusion from Logical Steps

Since a closed surface encloses a volume entirely without a boundary, and by applying Stokes' Theorem, the integral of the curl over a closed surface is zero. Logically, any vector field \(\mathbf{F}\) emanating from or swirling around tends to cancel out across the entire closed surface, resulting in a surface integral of zero.

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Key Concepts

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

Closed Surface
In the realm of vector calculus, a closed surface is one that completely encloses a three-dimensional volume. This is important because a closed surface has no edges or boundaries. Imagine a basketball; all the air is inside, and you cannot find an edge like you might on a piece of paper. Here's why it's essential:
  • A closed surface forms the complete boundary of a volume, like a bubble in air.
  • Common examples include spheres, cubes, and cylinders with top and bottom caps.
  • None of these surfaces have any openings or edges.
Thus, when it comes to calculus, a closed surface does not have a boundary line where you might calculate a line integral, which has crucial implications when applying calculus theorems such as Stokes' Theorem.
Surface Integral
Surface integrals extend the concept of integrals to functions over a surface in three-dimensional space. Instead of summing up along a line or over an area, you're summing up over a curved surface.
  • This kind of integral helps to determine quantities that are spread out over a surface, like flux across a membrane.
  • Mathematically, the surface integral of a vector field \(\mathbf{F}\) is expressed as \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\), integrating over the surface \(S\).
When you're dealing with closed surfaces, the surface integral often needs to consider properties like the curl of the vector field, which can lead to conclusions like those drawn from Stokes' Theorem where the integral can be zero due to symmetry or lack of a boundary.
Vector Field
Vector fields are mathematical structures where each point in space has a vector associated with it. This concept is crucial for understanding the behavior of fields like gravitational or electromagnetic fields in space.
  • A vector in this field might represent direction and magnitude, like wind speed and direction at different points.
  • Mathematically, a vector field \(\mathbf{F}\) is described using components \((P, Q, R)\) depending on the three-dimensional space coordinates \((x, y, z)\).
In calculus, particularly with Stokes' Theorem, notable operations involve the curl of applied vector fields. The curl runs through these vector functions like a twisting motion, determining rotational properties at a point鈥攁 crucial process when evaluating the behavior of vector fields on closed surfaces.
Line Integral
Line integrals are a fundamental calculus tool used to integrate functions over a curve. This differs from typical integrals in that it follows a path through space, which can be along a boundary or an open path.
  • A line integral of a vector field \(\mathbf{F}\) along a curve \(C\) is designated as \(\oint_C \mathbf{F} \cdot d \mathbf{r}\).
  • This measures things like the work done by a force field along a path.
In the context of closed surfaces, line integrals become particularly interesting. Because closed surfaces lack boundaries, the line integral turns out to be zero. Under Stokes' Theorem, this implies that when integrating around the boundary of a surface that doesn't physically exist 鈥� like on a completely enclosed surface 鈥� the result of this theoretical process simulates zero, leading to insightful realizations about the field interactions.

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Most popular questions from this chapter

Find \(\iint_{S}(\nabla \times \mathbf{F}) \cdot d \mathbf{S},\) where \(S\) is the ellipsoid \(x^{2}+y^{2}+2 z^{2}=10\) and \(\mathbf{F}\) is the vector field \(\mathbf{F}=(\sin x y) \mathbf{i}+e^{x} \mathbf{j}-y z \mathbf{k}\)

Use Gauss' law and symmetry to prove that the electric field due to a charge \(Q\) evenly spread over the surface of a sphere is the same outside the surface as the field from a point charge \(Q\) located at the center of the sphere. What is the field inside the sphere?

Integrate \(\nabla \times \mathbf{F}, \mathbf{F}=\left(3 y,-x z,-y z^{2}\right)\) over the portion of the surface \(2 z=x^{2}+y^{2}\) below the plane \(z=2,\) both directly and by using Stokes' theorem.

Show that the following vector fields are conservative. Calculate \(\int_{C} \mathbf{F} \cdot d \mathbf{s}\) for the given curve. (a) \(\quad F=\left(x y^{2}+3 x^{2} y\right) i+(x+y) x^{2} j ; C\) is the curve consisting of line segments from (1,1) to (0,2) to (3,0) (b) \(\mathbf{F}=\frac{2 x}{y^{2}+1} \mathbf{i}-\frac{2 y\left(x^{2}+1\right)}{\left(y^{2}+1\right)^{2}} \mathbf{j}: C\) is parametrized by \(x=t^{3}-1, y=t^{6}-t, 0 \leq t \leq 1\) (c) \(\mathbf{F}=\left[\cos \left(x y^{2}\right)-x y^{2} \sin \left(x y^{2}\right)\right] \mathbf{i}-2 x^{2} y \sin \left(x y^{2}\right) \mathbf{j}\) \(C\) is the curve \(\left(e^{t}, e^{t+1}\right),-1 \leq t \leq 0\)

Find the area bounded by one are of the cycloid \(x=a(\theta-\sin \theta), y=a(1-\cos \theta),\) where \(a>0,\) and \(0 \leq \theta \leq 2 \pi,\) and the \(x\) axis (use Green's theorem).
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