Calculating Curved Surface Area by Rotating Planes, Lecture notes of Differential and Integral Calculus

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FURTHER APPLICATIONS OF INTEGRATION

Area of a Surface

of Revolution

FURTHER APPLICATIONS OF INTEGRATION

AREA OF A SURFACE OF REVOLUTION We want to define the area of a surface of revolution in such a way that it corresponds to our intuition. ▪ If the surface area is A , we can imagine that painting the surface would require the same amount of paint as it does in a flat region with area A.

CIRCULAR CYLINDERS The lateral surface area of a circular cylinder with radius r and height h is taken to be: A = 2 πrh ▪ We can imagine cutting the cylinder and unrolling it to obtain a rectangle with dimensions of 2 πrh and h.

CIRCULAR CONES We know that, in general, the area of a sector of a circle with radius l and angle θ is ½ l 2 θ.

CIRCULAR CONES

So, the area is:

▪ Thus, we define the lateral surface area of a cone to be A = πrl. 1 2 1 2 2 2

2 r

A l l rl

l

  

Then, the approximating surface consists

of a number of bands—each formed by

rotating a line segment about an axis.

BANDS

BANDS

To find the surface area, each of these

bands can be considered a portion of

a circular cone.

BANDS From similar triangles, we have: This gives: 1 1 1 2 l l l r r

= 2 1 1 1 1 2 1 1 1 or ( ) = + − = r l r l r l r r l r l

BANDS Putting this in Equation 1, we get or where r = ½( r 1

• r 2 ) is the average radius of the band. 1 2

A =  ( r l + r l )

Formula 2

A = 2  rl

SURFACE AREA To define its surface area, we divide the interval [ a , b ] into n subintervals with endpoints x 0 , x 1 ,... , x n and equal width Δx as we did in determining arc length.

SURFACE AREA

If y

i

= f ( x

i

), then the point P

i

( x

i

, y

i

) lies

on the curve.

▪ The part of the surface between x i– 1 and x i is approximated by taking the line segment P i– 1 P i and rotating it about the x - axis.

SURFACE AREA As in the proof of Theorem 2 in Section 8.1, we have where x i

• is some number in [ x i – 1 , x i

].

  2 1

i i

P P f x x

−

SURFACE AREA When Δ x is small, we have y i = f ( x i ) ≈ f ( x i

and y i – 1 = f ( x i – 1 ) ≈ f ( x i * ), since f is continuous. Therefore, 2 1 *^ * 1 2 2 ( ) 1 '( ) 2 i i i i i i y y  P P  f x f x x − −

 + ^    