Explanation Given info: w Be the region below the parabolic x 2 + y 2 = 2 − z and lies above the part of plane x + y + z = 1 in the first octant. x ≥ 0 y ≥ 0 z ≥ 0 Concept used: Decide limit of z , draw the 3 − d diagram take projection of bounded region on x − y plane to decide the limit of variable x and y . Calculation: Form the above graph it becomes clear that z varies above x + y + z = 1 to x 2 + y 2 = z − 2 Hence, limit of z : 1 − x − y . ≤ z ≤ 2 + 2 + x 2 + y 2 ≥ 0 , y ≥ 0 , z ≥ 0 , so point must lies in the first octant. Take projection of bonded region on x y plane in horizontal plane bounded region looks like: x ≥ 0 , y ≥ 0 , z ≥ 0 , Projection on x y is bounded by x ≥ 0 , y ≥ 0 , and must be z ≥ 0. From projection you may decide limit of x and y 0 ≤ y ≤ 1 − x 0 ≤ x ≤ 1...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 16.3, Problem 26E

To determine

Let $w$ be the region below the parabolic $x^{2} + y^{2} = z - 2$

That lies above the part of the plane $x + y + z = 1$ in the first octant $(x \geq 0, y \geq 0) .$ express

$\underset{W}{∭} f (x, y, z) d v$ As an integrated integral $(for arbitrary function f)$

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Chapter 16.3, Problem 25E

Chapter 16.3, Problem 27E

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Chapter 16 Solutions

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Let w be the region below the parabolic x 2 + y 2 = z − 2 That lies above the part of the plane x + y + z = 1 in the first octant ( x ≥ 0 , y ≥ 0 ) . express ∭ W f ( x , y , z ) d v As an integrated integral ( for arbitrary function f )

Solution Summary: The author explains that w is the region below the parabolic x 2 + y 2 = z 2 and lies above the part of the plane

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Let $w$ be the region below the parabolic $x^{2} + y^{2} = z - 2$

That lies above the part of the plane $x + y + z = 1$ in the first octant $(x \geq 0, y \geq 0) .$ express

$\underset{W}{∭} f (x, y, z) d v$ As an integrated integral $(for arbitrary function f)$

Want to see the full answer?

Chapter 16 Solutions

Let w be the region below the parabolic x 2 + y 2 = z − 2 That lies above the part of the plane x + y + z = 1 in the first octant ( x ≥ 0 , y ≥ 0 ) . express ∭ W f ( x , y , z ) d v As an integrated integral ( for arbitrary function f )

Solution Summary: The author explains that w is the region below the parabolic x 2 + y 2 = z 2 and lies above the part of the plane

Videos

Let w be the region below the parabolic x2+y2=z−2 That lies above the part of the plane x+y+z=1 in the first octant (x≥0,y≥0). express ∭Wf(x,y,z)dv As an integrated integral (for arbitrary function f)

Want to see the full answer?

Chapter 16 Solutions

Let $w$ be the region below the parabolic $x^{2} + y^{2} = z - 2$

That lies above the part of the plane $x + y + z = 1$ in the first octant $(x \geq 0, y \geq 0) .$ express

$\underset{W}{∭} f (x, y, z) d v$ As an integrated integral $(for arbitrary function f)$