   Chapter 14.5, Problem 37E

Chapter
Section
Textbook Problem

Surface Area Find the surface area of the solid of intersection of the cylinders x 2 + z 2 = 1 and x 2 + z 2 = 1 (see figure). To determine

To calculate: The surface area of solid of intersection of area if the cylinders x2+z2=1 and y2+z2=1

Explanation

Given:

The given two cylinders are x2+z2=1 and y2+z2=1 as shown in the figure,

Formula used:

The surface area can be found by,

S=R1+[fx(x,y)]2+[fy(x,y)]2dA

The chain rule is:

ddxf(g(x))=f(g(x))g(x).

The differentiation formula is  ddx(xn)=nxn1, ddx(constant)=0.

Calculation:

The equation for the surface is,

z=1x2

First find the partial derivative with respect to x by the chain rule

ddxf(g(x))=f(g(x))g(x), ddx(xn)=nxn1, ddx(constant)=0.

The derivative is:

fx(x,y)=ddx(1x2)=(1x2)122(2x21)=x1x2

Againdifferentiate with respect to y is:

fx(x,y)=ddy(1x2)=0

Now put the above derivatives in the formula

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