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The integral is sometimes called the vector area of the

Introduction to Electrodynamics | 4th Edition | ISBN: 9780321856562 | Authors: David J. Griffiths ISBN: 9780321856562 45

Solution for problem 62P Chapter 1

Introduction to Electrodynamics | 4th Edition

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Introduction to Electrodynamics | 4th Edition | ISBN: 9780321856562 | Authors: David J. Griffiths

Introduction to Electrodynamics | 4th Edition

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Problem 62P

Problem 62P

The integral

is sometimes called the vector area of the surface S. If S happens to be flat, then |a| is the ordinary (scalar) area, obviously.

(a) Find the vector area of a hemispherical bowl of radius R.

(b) Show that a = 0 for any closed surface. [Hint: Use Prob. 1.61a.]

(c) Show that a is the same for all surfaces sharing the same boundary.

(d) Show that

Reference problem 1.61a

Although the gradient, divergence, and curl theorems are the fundamental integral theorems of vector calculus, it is possible to derive a number of corollaries from them. Show that:

where the integral is around the boundary line. [Hint: One way to do it is to draw the cone subtended by the loop at the origin. Divide the conical surface up into infinitesimal triangular wedges, each with vertex at the origin and opposite side dl, and exploit the geometrical interpretation of the cross product (Fig. 1.8).]

(e) Show that

for any constant vector c. [Hint: Let T = c · r in Prob. 1.61e.]

figure 1.8

Reference problem 1.61a

Step-by-Step Solution:

Solution

Step 1 of 6

We need to Find the vector area of a hemispherical bowl of radius .

The vector area of the surface s is    .

 is the area element spanning from  to  and  to on a spherical surface at constant radius .

The surface is considered to be in the northern hemisphere, the , ,  components are  

        

The  and  components integrate to zero and  component of   is .

So the vector area of the surface s is  .

                                 .

                                 

                                 

                                 

The vector area of a hemispherical bowl of radius  is found out to be .


Step 2 of 6

Chapter 1, Problem 62P is Solved
Step 3 of 6

Textbook: Introduction to Electrodynamics
Edition: 4
Author: David J. Griffiths
ISBN: 9780321856562

The answer to “The integral is sometimes called the vector area of the surface S. If S happens to be flat, then |a| is the ordinary (scalar) area, obviously.(a) Find the vector area of a hemispherical bowl of radius R.(b) Show that a = 0 for any closed surface. [Hint: Use Prob. 1.61a.](c) Show that a is the same for all surfaces sharing the same boundary.(d) Show that Reference problem 1.61aAlthough the gradient, divergence, and curl theorems are the fundamental integral theorems of vector calculus, it is possible to derive a number of corollaries from them. Show that: where the integral is around the boundary line. [Hint: One way to do it is to draw the cone subtended by the loop at the origin. Divide the conical surface up into infinitesimal triangular wedges, each with vertex at the origin and opposite side dl, and exploit the geometrical interpretation of the cross product (Fig. 1.8).](e) Show that for any constant vector c. [Hint: Let T = c · r in Prob. 1.61e.]figure 1.8 Reference problem 1.61a” is broken down into a number of easy to follow steps, and 172 words. Since the solution to 62P from 1 chapter was answered, more than 701 students have viewed the full step-by-step answer. This full solution covers the following key subjects: show, Vector, area, integral, surface. This expansive textbook survival guide covers 12 chapters, and 550 solutions. The full step-by-step solution to problem: 62P from chapter: 1 was answered by , our top Physics solution expert on 07/18/17, 05:41AM. This textbook survival guide was created for the textbook: Introduction to Electrodynamics , edition: 4. Introduction to Electrodynamics was written by and is associated to the ISBN: 9780321856562.

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