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Chapter 15.5, Problem 22E
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### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# If you attempt to use Formula 2 to find the area of the top half of the sphere x2 + y2 + z2 = a2, you have a slight problem because the double integral is improper. In fact, the integrand has an infinite discontinuity at every point of the boundary circle x2 + y2 = a2. However, the integral can be computed as the limit of the integral over the disk x2 + y2 ≤ t2 as t → a-. Use this method to show that the area of a sphere of radius a is 4πa2.

To determine

To show: The area of the sphere of radius a is equal to 4πa2 .

Explanation

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)

Given:

The sphere, x2+y2+z2=a2 .

Calculation:

Solve the given equation as shown below.

x2+y2+z2=a2z2=a2x2y2z=a2x2y2

The partial derivatives fx and fy are,

fx=2x2a2x2y2=xa2x2y2fy=2y2a2x2y2=ya2x2y2

Then, by the formula mentioned above, the area of the surface is given by,

A(S)=D(xa2x2y2)2+(ya2x2y2)2+1dA=Dx2a2x2y2+y2a2x2y2+1dA=Dx2+y2+a2x2y2a2x2y2dA=Da2a2x2y2dA

Use polar coordinates to solve

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