Find the surface area, in square units and correct to the nearest hundredth of a square unit, of the surface of revolution found by revolving the arc segment y = x^2 from (0, 0) to (3, 9), about the x-axis.
The radius would be x².
You're trying to find the circumference of a disk, so it's 2πx².
The slant height is √(1+(2x)²) = √(1+4x²)
∫2πx²√(1+4x²) dx
u = 1+4x²
du = 8x dx
dx = 1/(8x) du
x = √(u-1)/2
So
∫2πx²√(1+4x²) dx
2π∫x²√u 1/(8x)du
π/4∫x√u du
π/4∫√(u-1)/2√u du
π/8∫√(u-1)√u du
π/8∫√(u² - u) du
Then I guess you can do
v = u² - u
dv = 2u du
du = dv / 2u
π/16∫√v/v dv
π/16[2/3v^(3/2)]
π/16[2/3(u² - u)^(3/2)]
π/16[2/3((1+4x²)² - (1+4x²))^(3/2)]
evaluated from x=0 to x=3...
I probably screwed up something.
???
I think the answer is 208.09
a. 3.81
b. 49.42
c. 53.23
d. 208.09
e. 257.51
f. 261.32
g. none of these
2006-12-03
04:20:03
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2 answers
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asked by
Olivia
4
in
Mathematics