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how do you find the area of the surface obtained by rotating the curve about the x-axis? Given hint in the question: write $y$ in terms of $x$.

$3y^2 = x(1-x)^2,\ 0\leq x\leq 1$

I got $\frac{1}{3}\int_{0}^{1}\sqrt{12x(1-x)^2+(1-4x+3x^2)^2} \ dx$ and I'm stuck.

Any hints would be appreciated. Thanks!

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1 Answer 1

up vote 2 down vote accepted

Is there a problem with integral or what? If yes, then just open the brackets in it

$$9x^4 - 12x^3 - 2x^2 + 4x + 1 = (x-1)^2 (3x+1)^2$$ 2) now you have this:

$$\int(x-1)(3x+1)dx = \int(3x^2 - 2x + 1)dx = x^3 - x^2 + x$$ 3) and subtitute $0$ and $1$. Your integral $= 1$.

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Hey thanks for the answer. I got a question though, how do you factorize $9x^4 - 12x^3 - 2x^2 + 4x + 1$ into $(x-1)^2 * (3x+1)^2$? –  uohzxela Nov 13 '12 at 12:56
1  
Well, you see your constant term of the polynomial = 1. We search for its prime factors. It devides only by 1. So we substitute the value 1 or -1 into the equation. If the equation turns to 0 then our value is a root. The value = 1 is suitable, so we devide our equation by (x - value) = (x - 1) And then we have $$9x^4 − 12x^3 − 2x^2 + 4x + 1 = (x - 1)(9x^3 - 3x^2 - 5x - 1)$$ And then repeat these operations with the second brackets from the very beginning until you have the upper result. –  nenuka Nov 14 '12 at 6:47

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