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Rotate about the $x$ axis

$x = 2t-2/3t^3$

$y = 2t^2$

$0 \leq t \leq 1$

I did the integral of $\sqrt{(2-2t^2)^2+(4t)^2}$ and got $(2x(x^2+3))/3$ and then I did the integral of $2\pi 2t^2 ((2x(x^2+3))/3)$ and I get $22\pi/9$ but apparently that's not the correct answer

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I was right the first time: you appear to have integrated your infinitesimal arclength element (which is found correctly) before multiplying by $ \ y(t) \ $ . You also will find this easier to work with if you leave everything in parametric form when setting up the integral. –  RecklessReckoner Mar 27 at 17:57

1 Answer 1

The surface area integral for a figure revolved about the $ \ x-$ axis is

$$ S \ = \ 2 \pi \ \int \ y \ \ ds \ , $$

which, for a parametric curve, will be

$$ S \ = \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dx}{dt} \right)^2} \ \ dt \ \ = \ \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( 2 - 2t^2 \right)^2 + \left( 4t \right)^2} \ \ dt , $$

$$ = \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( 2 + 2t^2 \right)^2} \ \ dt \ \ = \ 2 \pi \ \int \ y(t) \ \cdot \ \underbrace{(2 + 2t^2) \ \ dt}_{ds} \ \ , $$

as you have set up. However, while your arclength has been determined correctly, you have to multiply the factors of the integrand first before you set about performing the integration itself. So you want

$$ S \ = \ 2 \pi \ \int_0^1 \ ( \ 2t^2 \ ) \ \cdot \ (2 + 2t^2) \ \ dt \ \ = \ \ \ 2 \pi \ \int_0^1 \ 4t^2 \ + \ 4t^4 \ \ dt $$

$$ = \ \ 2 \pi \ \left( \ \frac{4}{3}t^3 \ + \ \frac{4}{5}t^5 \right) \ \vert_0^1\ \ \ = \ \frac{64 \pi}{15} . $$

[corrected from what I did late last night]

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