Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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 $= 1$.

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.