Explanation Let there be a continuous function f ( x , y ) on R = [ a , b ] × [ c , d ] , then the below integrals are referred to as iterated integrals. ∬ R f ( x , y ) d A = ∫ a b ∫ c d f ( x , y ) d y d x = ∫ c d ∫ a b f ( x , y ) d x d y The double integral gives the volume of the box having base as R and height as 4. Given: The function is ∬ R ( 15 − 3 x ) d A and rectangle is given as R = [ 0 , 5 ] × [ 0 , 3 ] Formula: Length and Breadth of rectangle is given as Length, l = Δ x = x 2 − x 1 Breadth, b = Δ y = y 2 − y 1 Area of rectangle is given as Δ A = Δ x Δ y Calculation: The double integration is given as ∬ R ( 15 − 3 x ) d A = l ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.1, Problem 9E

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Evaluate the $\iint_{R} (15 - 3 x) d A$ , where $R = [0, 5] \times [0, 3]$ and also sketch the corresponding the solid region.

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Chapter 16.1, Problem 8E

Chapter 16.1, Problem 10E

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Chapter 16 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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Evaluate the ∬ R ( 15 − 3 x ) d A , where R = [ 0 , 5 ] × [ 0 , 3 ] and also sketch the corresponding the solid region.

Solution Summary: The author analyzes the function R over rectangle R = [ 0, 5 ] and sketches the corresponding solid region. The double integral gives the volume of the box having base as R.

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Evaluate the $\iint_{R} (15 - 3 x) d A$ , where $R = [0, 5] \times [0, 3]$ and also sketch the corresponding the solid region.

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Chapter 16 Solutions

Evaluate the ∬ R ( 15 − 3 x ) d A , where R = [ 0 , 5 ] × [ 0 , 3 ] and also sketch the corresponding the solid region.

Solution Summary: The author analyzes the function R over rectangle R = [ 0, 5 ] and sketches the corresponding solid region. The double integral gives the volume of the box having base as R.

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Evaluate the ∬R(15−3x) dA, where R = [0, 5] × [0, 3]and also sketch the corresponding the solid region.

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Chapter 16 Solutions

Evaluate the $\iint_{R} (15 - 3 x) d A$ , where $R = [0, 5] \times [0, 3]$ and also sketch the corresponding the solid region.