# Thread: Surface Area Problem - 3

1. ## Surface Area Problem - 3

Find the area of the surface by rotating the curve about the x axis

$y = \dfrac{1}{2}x^{3} + 6x^{-1}$

$\dfrac{1}{2} \leq x \leq 1$

$y' = \dfrac{3}{2}x^{2} - 6x^{-2}$

$S = \int_{1/2}^{1} 2\pi(y) \sqrt{1 + (\dfrac{3}{2}x^{2} - 6x^{-2})^{2}} dx$

Squared

$(\dfrac{3}{2}x^{2} - 6x^{-2})(\dfrac{3}{2}x^{2} - 6x^{-2}) = \dfrac{9}{2}x^{4} -18x + 36x^{-4}$

$S = 2\pi \int_{1/2}^{1} (\dfrac{1}{2}x^{3} + 6x^{-1}) \sqrt{1 + (\dfrac{9}{2}x^{4} -18x + 36x^{-4} )} dx$

Next move?

2. ## Re: Surface Area Problem - 3

\displaystyle \begin{align*} \left( \frac{3}{2}x^2 - 6x^{-2} \right) ^2 = \frac{9}{4}x^4 - 18 + 36x^{-4} \end{align*}, not \displaystyle \begin{align*} \frac{9}{4}x^4 - 18x + 36x^{-4} \end{align*}.

Once you have fixed this up in your integrand, get a common denominator and simplify.

3. ## Re: Surface Area Problem - 3

Find the area of the surface by rotating the curve about the x axis

$y = \dfrac{1}{2}x^{3} + 6x^{-1}$

$\dfrac{1}{2} \leq x \leq 1$

$y' = \dfrac{3}{2}x^{2} - 6x^{-2}$

$S = \int_{1/2}^{1} 2\pi(y) \sqrt{1 + (\dfrac{3}{2}x^{2} - 6x^{-2})^{2}} dx$

Squared

$(\dfrac{3}{2}x^{2} - 6x^{-2})(\dfrac{3}{2}x^{2} - 6x^{-2}) = \dfrac{9}{2}x^{4} -18 + 36x^{-4}$

$S = 2\pi \int_{1/2}^{1} (\dfrac{1}{2}x^{3} + 6x^{-1}) \sqrt{1 + \dfrac{9}{2}x^{4} -18 + 36x^{-4} )} dx$

You could multiply everything by 2 to get rid of the fraction

$S = 2\pi \int_{1/2}^{1} (\dfrac{1}{2}x^{3} + 6x^{-1}) \sqrt{2+ 18x^{4} -36 + 72x^{-4} )} dx$ ???

$S = 2\pi \int_{1/2}^{1} (\dfrac{1}{2}x^{3} + 6x^{-1}) \sqrt{18x^{4} -34 + 72x^{-4} )} dx$ ???