Nope. You only integrated 2πx √(1+4x^2)
Each strip of the surface has radius y=x^2, so
∫[0,4] 2πy ds
ds^2 = 1+y'^2 = 1+4x^2, so
A = ∫[0,4] 2π x^2 √(1+4x^2) dx
That's a bit more complicated:
http://www.wolframalpha.com/input/?i=%E2%88%AB[0%2C4]+%282%CF%80+x^2+%E2%88%9A%281%2B4x^2%29%29+dx
Compute the surface area of revolution about the x-axis over the interval.
y = x^2, [0, 4]
Is it (pi/12)(65^2 - 65)^(3/2)?
1 answer
1 answer