Calculus- Early Transcendentals, 2021a

334 Applications of Integration
is the surface area we seek. (Roughly speaking, this is because while x ∗ i and t i are distinct values in
[x i ,x i+1 ], they get closer and closer to each other as the length of the interval shrinks.)
.
(x i+1 , f (x i+1 ))
(x i , f (x i ))
x i x ∗ i x i+1
Figure 8.20: One subinterval.
Example 8.23: Surface Area of a Sphere
Compute the surface area of a sphere of radius r.
Solution. The sphere
√
can be obtained by rotating the graph of f (x) =
derivative f ′ is −x/ r 2 − x 2 , so the surface area is given by
√
∫ r √
A = 2π r 2 − x 2 1 + x2
−r
r 2 − x 2 dx
∫ r
= 2π
= 2π
−r
∫ r
−r
√
r 2 − x 2 √
rdx= 2πr
r 2
r 2 − x 2 dx
∫ r
−r
1dx = 4πr 2
√
r 2 − x 2 about the x-axis. The
If the curve is rotated around the y-axis, the formula is nearly identical, because the length of the line
segment we use to approximate a portion of the curve doesn’t change. Instead of the radius f (x ∗ i ), weuse
the new radius ¯x i =(x i + x i+1 )/2, and the surface area integral becomes
∫ b √
2πx 1 +(f ′ (x)) 2 dx.
a
♣
Example 8.24: Surface Around y-axis
Compute the area of the surface formed when f (x) =x 2 between 0 and 2 is rotated around the
y-axis.