Download Area of a Surface of Revolution

AP Calculus Lesson #56 Mr. Jeckovich
Area of a Surface of Revolution
Def: a frustrum of a right circular cone is formed by slicing a cone, where the slice is parallel to the base.
b
r1 + r2
 1 + f' x ® ¶ x
Surface Area of a Frustrum: S = 2 rL where r =
and L = 



2
a 
Definition of the Area of a Surface of Revolution
Let y = f x have a continuous derivative on the interval [a , b]. The area S of the surface of revolution formed
by revolving the graph of f about a horizontal or vertical axis is:
b
r x 1 + f' x ® ¶ x where r(x) is the distance between the graph of f and the axis of revolution.
S = 2 



a   
If x = gy on the interval [c , d], then the surface area is:
d
 r y 1 + g' y ® ¶ y where r(y) is the distance between the graph of g and the axis of revolution.
S = 2 



c   
Examples: Set up and evaluate the definite integral for the area of the surface generated by revolving the curve:
1. y = x , 1 , 4
about the x-axis
3
2
1
-1
1
2
3
4
5
-1
-2
-3
2. x = y® 1, 2
about the x-axis.
3
2
1
-1
1
2
3
4
5
-1
-2
-3
3. y = 4 - x®
0,2
about the y-axis.
4
3
2
1
-4
-3
-2
-1
1
2
3
4
-1
-2
-3
-4
4. Use a graphing utility to approximate the surface area of the solid of revolution:
y = ln x 1, e about the y-axis.
4
3
2
1
-4
-3
-2
-1
1
2
3
4
-1
-2
-3
-4
5. Find the formula for the surface area of a sphere.
Assign #56: p.413 #17
p.432 #7
p.442 #5, 31, 37