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?
Best Answer
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.