Explanation Given information: f ( x , y , z ) = g ( x ) h ( y ) k ( z ) , B = [ a , b ] × [ c , d ] × [ p , q ] . Formula used: d d x ( F ( x ) ) = f ( x ) ⇒ ∫ f ( x ) d x = F ( x ) Proof: Let d d x ( G ( x ) ) = g ( x ) ⇒ ∫ g ( x ) d x = G ( x ) , d d x ( H ( x ) ) = h ( x ) ⇒ ∫ h ( x ) d x = H ( x ) , and d d x ( K ( x ) ) = k ( x ) ⇒ ∫ k ( x ) d x = K ( x ) Now to prove ∭ B f ( x , y , z ) d V = ( ∫ a b g ( x ) d x ) ( ∫ c d g ( y ) d y ) ( ∫ p q g ( z ) d z ) First evaluate left-hand side, which can be converted in iterated form as ∭ B f ( x , y , z ) d V = ∫ a b ∫ c d ∫ p q g ( x ) h ( y ) k ( z ) d z d y d x To evaluate this first evaluate inner integral treating functions of x and y constant, thus ∫ p q g ( x ) h ( y ) k ( z ) d z = g ( x ) h ( y ) ∫ p q k ( z ) d z = g ( x ) h ( y ) [ K ( z ) ] p q = g ( x ) h ( y ) [ K ( q ) − K ( p ) ] Now, integrate above with respect to y treating x as constant, thus ∫ c d g ( x ) h ( y ) [ K ( q ) − K ( p ) ] d y = g ( x ) [

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ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Chapter 15.3, Problem 27E

To determine

Show that the integral of f over box B=[a,b]×[c,d]×[p,q], where f(x,y,z)=g(x)h(y)k(z) can be expressed as a product of integral as follow:

∭Bf(x,y,z)dV=(∫abg(x)dx)(∫cdg(y)dy)(∫pqg(z)dz)

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Chapter 15.3, Problem 26E

Chapter 15.3, Problem 28E

Chapter 15 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

Ch. 15.1 - Prob. 1PQ Ch. 15.1 - Prob. 2PQ Ch. 15.1 - Prob. 3PQ Ch. 15.1 - Prob. 4PQ Ch. 15.1 - Prob. 5PQ Ch. 15.1 - Prob. 6PQ Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E

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Show that the integral of f over box B = [ a , b ] × [ c , d ] × [ p , q ] , where f ( x , y , z ) = g ( x ) h ( y ) k ( z ) can be expressed as a product of integral as follow: ∭ B f ( x , y , z ) d V = ( ∫ a b g ( x ) d x ) ( ∫ c d g ( y ) d y ) ( ∫ p q g ( z ) d z )

Solution Summary: The author explains that the integral of f over box B=left[a,bright]times left[c,d right]

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Show that the integral of f over box B=[a,b]×[c,d]×[p,q], where f(x,y,z)=g(x)h(y)k(z) can be expressed as a product of integral as follow:

∭Bf(x,y,z)dV=(∫abg(x)dx)(∫cdg(y)dy)(∫pqg(z)dz)

Want to see the full answer?

Chapter 15 Solutions

Show that the integral of f over box B = [ a , b ] × [ c , d ] × [ p , q ] , where f ( x , y , z ) = g ( x ) h ( y ) k ( z ) can be expressed as a product of integral as follow: ∭ B f ( x , y , z ) d V = ( ∫ a b g ( x ) d x ) ( ∫ c d g ( y ) d y ) ( ∫ p q g ( z ) d z )

Solution Summary: The author explains that the integral of f over box B=left[a,bright]times left[c,d right]

Videos

Show that the integral of f over box B=[a,b]×[c,d]×[p,q], where f(x,y,z)=g(x)h(y)k(z) can be expressed as a product of integral as follow: ∭Bf(x,y,z)dV=(∫abg(x)dx)(∫cdg(y)dy)(∫pqg(z)dz)

Want to see the full answer?

Chapter 15 Solutions

Show that the integral of f over box B=[a,b]×[c,d]×[p,q], where f(x,y,z)=g(x)h(y)k(z) can be expressed as a product of integral as follow:

∭Bf(x,y,z)dV=(∫abg(x)dx)(∫cdg(y)dy)(∫pqg(z)dz)