   Chapter 15.5, Problem 5E

Chapter
Section
Textbook Problem

Find the area of the surface.5. The part of the paraboloid z = 1 – x2 – y2 that lies above the plane z = –2.

To determine

To find: The area of given surface.

Explanation

Given:

The part of the paraboloid, z=1x2y2 .

The region D lies above the plane z=2 .

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:

Solve the given equations,

z=1x2y22=1x2y2x2y2=3x2+y2=3

Since it is the equation of the circle of radius 3 , it is more appropriate to consider the polar coordinates. Hence, r varies from 0 to 3 and θ varies from 0 to 2π .

The partial derivatives fx and fy are,

fx=2xfy=2y

Then, obtain the area of surface by using equation (1).

A(S)=D(2x)2+(2y)2+1dA=02π034(x2+y2)+1dydx=02π034(r2)+1(r)drdθ=02π03r4r2+1drdθ

Use the equation (2) to separate the integrals by substituting t=4r2,dt=8rdr

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