Explanation Given: The region is f ( x , y ) = 16 − x 2 − y 2 . Formula Used: The surface area can be calculated of the region R by, S = ∬ R 1 + [ f x ( x , y ) ] 2 + [ f y ( x , y ) ] 2 d A Apply power rule of integration , ∫ x n d x = x n + 1 n + 1 . Differentiation formula d d x ( x n ) = n x n − 1 , d d x ( constant ) = 0 . The equation of circle x 2 + y 2 = r 2 , where r is the radius. Calculation: The function given is f ( x , y ) = 16 − x 2 − y 2 . Now partially differentiating it with respect to x , use d d x ( x n ) = n x n − 1 , d d x ( constant ) = 0 . f x ( x , y ) = d d x ( 16 − x 2 − y 2 ) = − 2 x 2 − 1 + 0 + 0 = − 2 x Now, with respect to y . It is found that, f y ( x , y ) = d d y ( 16 − x 2 − y 2 ) = 0 − 2 y 2 − 1 + 0 = − 2 y Substitute in the formula S = ∬ R 1 + [ f x ( x , y ) ] 2 + [ f y ( x , y ) ] 2 d A . S = ∬ R 1 + [ 2 x ] 2 + [ − 2 y ] 2 d A = ∬ R 1 + 4 x 2 + 4 y 2 d A = ∬ R 1 + 4 x 2 + 4 y 2 d A For provided region, the limits shall be converted to polar points for both x and y . Then, polar curve, x = r cos θ and y = r sin θ , which implies to, x 2 + y 2 = r 2 ( cos 2 x + sin 2 x ) = r 2 = 16 Since the region is in first octant so the obtained region is, Then the region is bounded by 0 ≤ r ≤ 4 and 0 ≤ θ ≤ π 4 Thus, on applying limits to the integral, using formula ∫ x n d x = x n + 1 n + 1 S = ∫ 0 π 2 ∫ 0 4 1 + 4 r 2 r d r d θ On substituting, u = ( 1 + 4 r 2 ) , d u = 8 r d r . Then, r d r = d u 8 . Calculate the integration of ∫ 0 4 1 + 4 r 2 r d r first

Bundle: Calculus: Early Transcendental Functions, 7th + Webassign, Multi-term Printed Access Card

7th Edition

ISBN: 9781337888936

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 14.5, Problem 18E

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To calculate: The area of the surface given by f(x,y)=16−x2−y2, in the first octant of the paraboloid.

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Chapter 14.5, Problem 17E

Chapter 14.5, Problem 19E

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Bundle: Calculus: Early Transcendental Functions, 7th + Webassign, Multi-term Printed Access Card

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The area of the surface given by f ( x , y ) = 16 − x 2 − y 2 , in the first octant of the paraboloid.

Solution Summary: The author explains how the surface area can be calculated of the region R by applying power rule of integration.

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To calculate: The area of the surface given by f(x,y)=16−x2−y2, in the first octant of the paraboloid.

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Chapter 14 Solutions