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p595, number 29.
Find the surface area of the ellipse
x
2
a
2
+
y
2
b
2
= 1
for
a > b
when rotated about
x
axis.
——————————————
Notice that this ellipse has the points (

a,
0) and (+
a,
0) as its
x
intercepts
and (

b,
0), (+
b,
0) as
y
intercepts. Check this.
So solve for
y
(
x
) and plug into the formula:
S.A.
=
Z
2
πf
(
x
)
ds
(1)
=
Z
2
πf
(
x
)
p
1 + (
f
0
(
x
))
2
dx
(2)
And then let
x
go from

a
to
a
.
Solve for
y
:
y
(
x
) =
bsqrt
1

(
x
a
)
2
.
Diﬀerentiate with respect to
x
y
0
(
x
) =

b
a
2
x
1
p
1

(
x
a
)
2
.
Plug into the formula:
S.A.
=
Z
2
πy
(
x
)
p
1 + (
y
0
(
x
))
2
dx
(3)
=
Z
a

a
2
πb
r
1

(
x
a
)
2
r
1 +
b
2
a
4
x
2
(1

(
x
a
)
2
)

1
dx.
(4)
All we need to do now is compute the integral. It’s a pain. There will be
lots of
u
sub. and notice how the bounds on the integral keep changin’ each
time. It’s okay to not keep track as long as we go back to
x
at the end.
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This note was uploaded on 07/11/2009 for the course MATH Math 31B taught by Professor Houdayer during the Winter '09 term at UCLA.
 Winter '09
 HOUDAYER
 Math, YIntercept

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