Math 1B
§ 8.2 Area of a Surface of Revolution
Overview: In this section we will find the surface area of a solid of revolution.
Let’s revolve the continuous function y = f ( x ) in the interval [a,b] about the x-axis.
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We’ll start by dividing the interval into n equal
subintervals of width △x. On each subinterval we
will approximate the function with a straight line
that agrees with the function at the endpoints of
each interval.
Now, revolve the approximations about the x-axis
and we get the following solid.
Surface Area Formulas:
S=
∫ 2π y ds
rotation about the x-axis
S=
∫ 2π x ds
rotation about the y-axis
2
where,
! dy $
ds = 1+ # & dx
" dx %
2
! dx $
ds = 1+ # & €dy
" dy %
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if y = f ( x ) , a ≤ x ≤ b
if €
x = g( y ) , c ≤ y ≤ d
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Think of 2π y or 2π x as the circumference of the “band” that is generated when revolving a section of
the curve of length ds around the x-axis or y-axis, respectively.
Example: Find the surface area of the portion of the sphere generated by revolving the curve
1
y = 1− x 2 , 0 ≤ x ≤ , about the x-axis.
2
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Example: Find the surface area generated by revolving the curve x = 1+ 2y 2 , 1 ≤ y ≤ 2 , about the xaxis.
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Example: Find the surface area generated by revolving the curve y = 3 3x , 0 ≤ y ≤ 2 , about the y-axis.
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Example: Set up an integral for the surface area generated by revolving y = x 2 , 0 ≤ x ≤ 2 , about the
a) x-axis
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b) y-axis
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