   Chapter 15.5, Problem 9E

Chapter
Section
Textbook Problem

Find the area of the surface.9. The part of the surface z = xy that lies within the cylinder x2 + y2 = 1

To determine

To find: The area of given surface.

Explanation

Given:

The part of the surface z=xy.

The region D lies within the cylinder x2+y2=1.

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.

If f is a polar rectangle R given by 0arb,αθβ, where 0βα2π then,Rf(x,y)dA=αβabf(rcosθ,rsinθ)rdrdθ (1)

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (2)

Calculation:

Convert the given problems into the polar coordinates to make the problem easier. By given the conditions, it is observed that, r varies from 0 to 1 and θ varies from 0 to 2π.

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

fx=yfy=x

Then, by the equation (1), the area of surface is given by,

A(S)=D(y)2+(x)2+1dA=02π01x2+y2+1dydx=02π01(r2)+1(r)drdθ=02π01rr2+1drdθ

Use the equation (2) to separate the integrals and substitute t=r2,dt=2rdr

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