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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Find the area of the surface.

8. The surface z = 2 3 ( x 3 2 + y 3 2 ) , 0 x 1 , 0 y 1

To determine

To find: The area of given surface.

Explanation

Given:

The function is the surface, z=23(x32+y32).

The limit variation of the region D is 0x1,0y1.

Formula used:

The surface area with equation z=f(x,y),(x,y)D, where fx and fy are continuous, is A(S)=D[fx(x,y)]2+[fy(x,y)]2+1dA.

Here, D is the given region.

Calculation:

Obtain the partial derivatives of f with respect to x and y.

fx=23(32)x12=x12fy=23(32)y12=y12

Then, the area of surface is given by,

A(S)=D(x12)2+(y12)2+1dA=0101x+y+1dydx

Integrate with respect to y and apply the limit of it.

A(S)=2301[(x+y+1)32]01dx=2301[(x+1+1)32(x+0+1)32]dx=2301[(x+2)32(x+1)32]dx

Integrate with respect to x and apply the limit of it

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