   Chapter 15, Problem 4P

Chapter
Section
Textbook Problem

If a, b, and c are constant vectors, r is the position vector x i + y j + z k, and E is given by the inequalities 0 ≤ a · r ≤ α, 0 ≤ b · r ≤ β, 0 ≤ c · r ≤ γ, show that ∭ E ( a ⋅  r ) ( b   ⋅   r ) ( c ⋅ r )   d V = ( α β γ ) 2 8 | a ⋅ ( b   ×   c ) |

To determine

To show: E(a.r)(b.r)(c.r)dV=(αβγ)28|a.(b×c)|.

Explanation

Given:

The constant vectors are denoted by a,b and c.

The position vector xi+yj+zk is denoted by r.

The region E is 0a.rα,0b.rβ,0c.rγ.

Property used: Change of Variable

Change of Variable in double integral is given by, Rf(x,y)dA=Sf(x(u,v),y(u,v))|(x,y)(u,v)|dudv (1)

Calculation:

Use the transformation u=a.r, v=b.r and w=c.r. Thus, the region E can be rewritten as 0uα,0vβ,0wγ.

Obtain the Jacobian by the definition of the dot and cross product.

(u,v,w)(x,y,z)=|xuxvxwyuyvywzuzuzu|=||a1a2a3b1b2b3c1c2c3||=|a.(b×c)|

By using the equation (1) the given integral will be,

RxydA=022014(u2v2)|12|dudv=180220(u2v2)dudv

Integrate the above integral with respect to u and apply the limit.

E(a

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