   Chapter 15.5, Problem 8E

Chapter
Section
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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