Explanation Given: f ( x , y , z ) = x and domain “D” is in the first octant bounded above by z = 8 − 2 x 2 − y 2 and below by z = y 2 . Formula used: The integral formula ∫ x n d x = x n + 1 n + 1 Calculation: The region is z-simple, so ∫ ∫ ∫ D x d v = ∫ ∫ w ∫ y 2 8 − 2 x 2 − y 2 x d z d A Now to evaluate the integral we must find the equation of the curved part of the boundary of w. Now, To find the boundary of w. The upper and lower surfaces intersect at points ( x , y , z ) where both z = 8 − 2 x 2 − y 2 and z = y 2 are satisfied. Thus 8 − 2 x 2 − y 2 or x 2 + y 2 = 4 The circle x 2 + y 2 = 4 in the first octant hits the axis at ( 2 , 0 ) and ( 0 , 2 ) (by putting y = 0 and x = 0 ) Since, domain is both vertical and horizontally simple, we can integrate in either dydx order or dydx. If we choose dydx then y varies from 0 to 2 and the domain is described by w; 0 ≤ y ≤ 2 , 0 ≤ x 4 − y 2 . Thus, the triple integral can be written as an iterated integral ∫ ∫ ∫ D x d v = ∫ 0 2 ∫ 0 4 − y 2 ∫ y 2 8 − 2 x 2 − y 2 x d z d x d y To solve the triple integral first calculate the inner integral first calculate the inner integral treating x and y as constant ∫ 0 2 ∫ 0 4 − y 2 ∫ y 2 8 − 2 x 2 − y 2 x d z d x d y = ∫ 0 2 ∫ 0 4 − y 2 x [ z ] y 2 8 − 2 x 2 − y 2 d x d y = ∫ 0 2 ∫ 0 4 − y 2 x ( 8 − 2 x 2 − y 2 − y 2 ) d x d y</

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ISBN: 9781319309671

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 15.3, Problem 17E

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Integrate f(x,y,z)=x over the region in the first octant bounded above by z=8−2x2−y2 and below by z=y2.

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Chapter 15.3, Problem 16E

Chapter 15.3, Problem 18E

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Chapter 15 Solutions

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS

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Integrate f ( x , y , z ) = x over the region in the first octant bounded above by z = 8 − 2 x 2 − y 2 and below by z = y 2 .

Solution Summary: The author explains the integral formula displaystyle, which is based on the curved part of the boundary of w.

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Integrate f(x,y,z)=x over the region in the first octant bounded above by z=8−2x2−y2 and below by z=y2.

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Chapter 15 Solutions