What is the surface area of the solid created by revolving #f(t) = ( t^3-3t+4, t^3-2t, t in [2,3]# around the x-axis?

Question

William Abdalla · Accepted Answer

See the answer below:

Isabella Ackerman · Answer

undefined To find the surface area of the solid created by revolving the curve $ f(t) = (t^3 - 3t + 4, t^3 - 2t) $ on the interval $ t $ in $[2,3]$ around the x-axis, you can use the formula for the surface area of a solid of revolution.

The surface area $ S $ is given by the formula:
$$ S = 2\pi \int_{a}^{b} y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt $$

Where $ a $ and $ b $ are the limits of integration.

First, find $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $:
$$ \frac{dx}{dt} = \frac{d}{dt}(t^3 - 3t + 4) = 3t^2 - 3 $$
$$ \frac{dy}{dt} = \frac{d}{dt}(t^3 - 2t) = 3t^2 - 2 $$

Now, compute $ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} $:
$$ \sqrt{\left(3t^2 - 3\right)^2 + \left(3t^2 - 2\right)^2} $$
$$ = \sqrt{9t^4 - 18t^2 + 9 + 9t^4 - 12t^2 + 4} $$
$$ = \sqrt{18t^4 - 30t^2 + 13} $$

Now, integrate $ y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} $ from $ t = 2 $ to $ t = 3 $:
$$ S = 2\pi \int_{2}^{3} (t^3 - 2t) \sqrt{18t^4 - 30t^2 + 13} dt $$

You can proceed to compute this integral to find the surface area of the solid of revolution.