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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

  1. 1. Let S be a smooth parametric surface and let P be a point such that each line that starts at P intersects S at most once. The solid angle Ω(S) subtended by S at P is the set of lines starting at P and passing through S. Let S(a) be the intersection of Ω(S) with the surface of the sphere with center P and radius a. Then the measure of the solid angle (in steradians) is defined to be

    |Ω(S)| = area of S ( a ) a 2

    Apply the Divergence Theorem to the part of Ω(S) between S(a) and S to show that

    |Ω(S)| = S r · n r 3 dS

    where r is the radius vector from P to any point on S, r = | r |, and the unit normal vector n is directed away from P.

    This shows that the definition of the measure of a solid angle is independent of the radius a of the sphere. Thus the measure of the solid angle is equal to the area subtended on a unit sphere. (Note the analogy with the definition of radian measure.) The total solid angle subtended by a sphere at its center is thus 4π steradians.

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To determine

To show: The value of |Ω(S)|=Srnr3dS .

Explanation

Given data:

|Ω(S)|=areaofS(a)a2

Formula used:

Write the expression for Divergence Theorem.

EdivFdV=SFdS (1)

Consider the portion of Ω(S) and S is represented by S1 and S1 is the boundary.

Also consider the lateral surface of S1 is represented by SL .

Modify equation (1) as follows.

S1rnr3dS=S1rr3dV (2)

Consider r in terms of x , y and z .

r=xi+yj+zk

Consider r in terms of x , y and z represented as follows.

r=x2+y2+z2

Find the value of rr3 .

rr3=x,y,zx(x2+y2+z2)32,y(x2+y2+z2)32,z(x2+y2+z2)32=x[x(x2+y2+z2)32]+y[y(x2+y2+z2)32]+z[z(x2+y2+z2)32]=(x2+y2+z23x2)+(x2+y2+z23y2)+(x2+y2+z23z2)(x2+y2+z2)52=0

Substitute 0 for rr3 in equation (2),

S1rnr3dS=S1(0)dV=0

The surface of S1 other than S(a) and S is results in rn=0 .

Find the value of SLrnr3dS .

SLrnr3dS=SL0r3dS=0

Expand S1rnr3dS in terms of S(a) , S and SL as follows

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Chapter 16 Solutions

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