Chapter 14.5, Problem 37E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
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=1Â andÂ y2+z2=1 as shown in the figure,

Formula used:

The surface area can be found by,

S=âˆ¬R1+[fx(x,y)]2+[fy(x,y)]2dA

The chain rule is:

ddxf(g(x))=fâ€²(g(x))â‹…gâ€²(x).

The differentiation formula is Â ddx(xn)=nxnâˆ’1,Â ddx(constant)=0.

Calculation:

The equation for the surface is,

z=1âˆ’x2

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

ddxf(g(x))=fâ€²(g(x))â‹…gâ€²(x),Â ddx(xn)=nxnâˆ’1,Â ddx(constant)=0.

The derivative is:

fx(x,y)=ddx(1âˆ’x2)=(1âˆ’x2)1âˆ’22â‹…(âˆ’2x2âˆ’1)=âˆ’x1âˆ’x2

Againdifferentiate with respect to y is:

fx(x,y)=ddy(1âˆ’x2)=0

Now put the above derivatives in the formula

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