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Set up an integral that represents the area of the surface obtained by rotating the given curve about the x-axis. Then use your calculator to find the surface area correct to four decimal places.

$$ x=t^2 - t^3, \qquad y= t + t^4 $$

on the interval $[0,1]$.

So I found $$ dx/dt = 2t - 3t^2 \quad\text{and}\quad dy/dt = 1 + 4t^3, $$ then I set up my integral as: $$ \int_0^1 2\pi (t + t^4 ) \sqrt{ (2t - 3t^2 )^2 + (1 + 4t^3 )^2 } \; dt $$ and got the answer $12.7176$.

Could anyone tell me if this is correct?

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Your integrand is right; I assume you evaluated the integral (probably with calculator or Wolfram0 correctly. – Chris K Dec 5 '13 at 1:46

You did $dx/dt = 2t - 3t^2$ and $dy/dt = 1 + 4t^3$ which leads us to have $$\frac{dy}{dx}=\frac{1 + 4t^3}{2t - 3t^2}$$ Now you were using this formula: $$P=\int_a^b2\pi y \sqrt{1+y'^2}dy$$ But what is $\sqrt{y'^2+1}$. $$\sqrt{y'^2+1}=\sqrt{1+\left(\frac{1 + 4t^3}{2t - 3t^2}\right)^2}=\frac{\sqrt{(2t-3t^2)^2+(1+4t^3)^2}}{\color{red}{|2t-3t^2|}}$$ Now you should examine if what is the sign of the denominator on $[0,1]$. Overall, your integral is correct and it is waiting to be solved by a proper method.

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