Explanation Given: The cone is z 2 = x 2 + y 2 where z ≥ 0 contained within the cylinder y 2 + z 2 ≤ 1 . Calculation: It is given that cone is z 2 = x 2 + y 2 . So; z = ± x 2 + y 2 It is also provided that the region is above x y -plane. So, consider only positive value of z that is; z = x 2 + y 2 Now, differentiate the z partially with respect to x and y as; ∂ z ∂ x = x x 2 + y 2 ∂ z ∂ y = y x 2 + y 2 To find the intersection part of the cone and cylinder substitute z 2 = 1 − y 2 in the cone equation x 2 + y 2 = z 2 as; x 2 + y 2 = 1 − y 2 x 2 + 2 y 2 = 1 The equation x 2 + 2 y 2 = 1 is an equation of ellipse. The figure is shown below: So; y 2 = 1 − x 2 2 y = ± 1 − x 2 2 So, y varies from − 1 − x 2 2 to 1 − x 2 2 . When y = 0 ; 1 − x 2 2 = 0 x = ± 1 So, x varies from − 1 to 1. Now, area can be calculated as; S = ∬ D 1 + ( ∂ z ∂ x ) 2 + ( ∂ z ∂ y ) 2 d A = ∬ D 1 + ( x x 2 + y 2 ) 2 + ( y x 2 + y 2 ) 2 d A = ∬ ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.4, Problem 37E

To determine

The surface area of the portion $S$ of the cone $z^{2} = x^{2} + y^{2}$ where $z \geq 0$ contained within the cylinder $y^{2} + z^{2} \leq 1$ .

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Chapter 17.4, Problem 36E

Chapter 17.4, Problem 38E

Chapter 17 Solutions

CALCULUS (CLOTH)

Ch. 17.1 - Prob. 1PQ Ch. 17.1 - Prob. 2PQ Ch. 17.1 - Prob. 3PQ Ch. 17.1 - Prob. 4PQ Ch. 17.1 - Prob. 1E Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E

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The surface area of the portion S of the cone z 2 = x 2 + y 2 where z ≥ 0 contained within the cylinder y 2 + z 2 ≤ 1 .

Solution Summary: The author calculates the surface area of the cone z 2 = x 2 + y 2 and determines its intersection with the cylinder.

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The surface area of the portion $S$ of the cone $z^{2} = x^{2} + y^{2}$ where $z \geq 0$ contained within the cylinder $y^{2} + z^{2} \leq 1$ .

Want to see the full answer?

Chapter 17 Solutions

The surface area of the portion S of the cone z 2 = x 2 + y 2 where z ≥ 0 contained within the cylinder y 2 + z 2 ≤ 1 .

Solution Summary: The author calculates the surface area of the cone z 2 = x 2 + y 2 and determines its intersection with the cylinder.

Videos

The surface area of the portion S of the cone z2=x2+y2 where z≥0 contained within the cylinder y2+z2≤1.

Want to see the full answer?

Chapter 17 Solutions

The surface area of the portion $S$ of the cone $z^{2} = x^{2} + y^{2}$ where $z \geq 0$ contained within the cylinder $y^{2} + z^{2} \leq 1$ .