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**Unformatted text preview: **e the integral.
4 L= = 1+ 2 9x 3 + 4 4 9x 0 2
3 dx = 4 2 9x 3 + 4 0 9x 2
3 4 dx = 0 2 9x 3 + 4
3x 1
3 2 dx = 4
0 1 −1
2
x 3 9x 3 + 4 dx
3 1 We now substitute using u = 9x 3 + 4 and du = 6x− 3 dx
This gives 1
6 2 9(4 3 )+4
4 2 11
13 3
u 2 du =
u2
3
18 2 1
Evaluating this gives:
27 94 2
3 +4 9(4 3 )+4
4
3
2 −8 2. (5 points) Find the area of the surface generated by rotating the curve y = x3 between x = 0 and
x = 2 about the x-axis.
b Recall that the basic formula for ﬁnding surface area in this situation is: SA = 2πf (x) 1 + [f ′ (x)]2 dx a
2 Here, f (x) = 3x , so we get:
′ SA = 2π 2
0 x3 1 + [3x2 ]2 dx = 2π 4
0 √
x3 1 + 9x4 dx We continue by using the substitution: u = 1 + 9x4 and du = 36x3 dx
This gives us the integral: 2π 145
1 π
11
u 2 du =
36
18 145
1 π 2 3
u2
u du =
18 3
1
2 145
1 3
π
145 2 − 1
=
27
Extra Credit: Derive a formula for the surface area of a sphere of radius r. [Give your work on the
back of the quiz or attach additional work]
√
Hint: Think of a sphere as being generated by revolving the semicircle given by f (x) = r2 − x2
about the x-axis and use a surface integral. The algebra is a little messy, but it works out....

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