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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Find the area of the surface.

11. The part of the sphere x2 + y2- z2 = a2 that lies within the cylinder x2 + y2 = ax and above the xy-plane

To determine

To find: The area of given surface.

Explanation

Given:

The function is the part of the sphere, x2+y2+z2=a2.

The region D lies within the cylinder x2+y2=ax above the xy plane.

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)

Calculation:

Solve the given equation of the function.

x2+y2+z2=a2z2=a2x2y2z=a2x2y2

Convert the given problems into the polar coordinates to make the problem easier. To obtain the limits, solve the given equations as shown below.

x2+y2=axr2=arcosθr=acosθ

By given the conditions, it is observed that, r varies from 0 to acosθ and θ varies from π2to π2.

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

fx=12(a2x2y2)12(2x)=x(a2x2y2)12fy=12(a2x2y2)12(2y)=y(a2x2y2)12

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

A(S)=D(x(a2x2y2)12)2+(y(a2x2y2)12)2+1dA=π2π20acosθx2(a2x2y2)+y2(a2x2y2)+1dydx=π2π20acosθr2+a2r2a2r2(r)drdθ=π2π20acosθra2a2r2drdθ=π2π20acosθara2r2drdθ

Substitute t=r2,dt=2rdr to integrate with respect to r

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Chapter 15 Solutions

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