# Thread: Surface Area Problem - 4

1. ## Surface Area Problem - 4

Find the surface area, considering the curve is being rotated about the y axis.

$\displaystyle y =$ 3rd root of $\displaystyle x$

$\displaystyle y = x^{1/3}$

$\displaystyle y^{3} = x$

$\displaystyle x = y^{3}$

$\displaystyle 2 \leq y \leq 3$

$\displaystyle x' = 3y^{2}$

$\displaystyle S = \int_{2}^{3} 2\pi(x) \sqrt{1 + (3y^{2})^{2}} dy$

$\displaystyle S = 2\pi\int_{2}^{3} (y^{3}) \sqrt{1 + (9y^{4})} dy$

Next move, probably trig?

2. ## Re: Surface Area Problem - 4

A little bit of simplification first, then yes, a trigonometric substitution

\displaystyle \begin{align*} S &= 2\pi \int_2^3{ y^3 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, \mathrm{d}y } \\ &= \frac{\pi}{3} \int_2^3{ y^2 \, \sqrt{ 1 + \left( 3y ^2 \right) ^2 } \, 6y\, \mathrm{d}y } \\ &= \frac{\pi}{9} \int_2^3{ 3y^2 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, 6y \, \mathrm{d}y } \end{align*}

So now substitute \displaystyle \begin{align*} 3y^2 = \tan{ \left( \theta \right) } \implies 6y \, \mathrm{d}y = \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta \end{align*} and notice that when \displaystyle \begin{align*} y = 2, \theta = \arctan{ \left( 12 \right) } \end{align*} and when \displaystyle \begin{align*} y = 3, \theta = \arctan{ \left( 27 \right) } \end{align*}, the integral becomes

\displaystyle \begin{align*} \frac{\pi}{9} \int_2^3{ 3y^2 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, 6y\, \mathrm{d}y } &= \frac{\pi}{9} \int_{\arctan{ \left( 12 \right) } } ^{\arctan{ \left( 27 \right) } } { \tan{ \left( \theta \right) } \, \sqrt{ 1 + \tan^2{ \left( \theta \right) } } \, \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta } \end{align*}

Go from here.

3. ## Re: Surface Area Problem - 4

Do you have to use trig sub stuff on this one?

4. ## Re: Surface Area Problem - 4

Find the surface area, considering the curve is being rotated about the y axis.

$\displaystyle y =$ 3rd root of $\displaystyle x$

$\displaystyle y = x^{1/3}$

$\displaystyle y^{3} = x$

$\displaystyle x = y^{3}$

$\displaystyle 2 \leq y \leq 3$

$\displaystyle x' = 3y^{2}$

$\displaystyle S = \int_{2}^{3} 2\pi(x) \sqrt{1 + (3y^{2})^{2}} dy$

$\displaystyle S = 2\pi\int_{2}^{3} (y^{3}) \sqrt{1 + (9y^{4})} dy$

Let's not use trig

$\displaystyle S = 2\pi\int_{2}^{3} (y^{3}) {1 + (9y^{4})^{1/2} dy$

$\displaystyle S = 2\pi\int_{2}^{3} (y^{3}) {u)^{1/2} dy$

$\displaystyle u = 1 + 9y^{4}$

$\displaystyle du = 36y^{3} dy$

$\displaystyle \dfrac{1}{36}du = y^{3} dy$

$\displaystyle S = 2\pi\int_{2}^{3} \dfrac{1}{36} (u)^{1/2} dy$ ?? On the right track here?

$\displaystyle = 2\pi \dfrac{1}{36}\dfrac{2}{3}(u)^{3/2}$

$\displaystyle = 2\pi \dfrac{1}{54}(u)^{3/2}$ evaluated at upper bound 3 and lower bound 2

$\displaystyle = 2\pi \dfrac{1}{54}(1 + 9y^{4})^{3/2}$ evaluated at upper bound 3 and lower bound 2

$\displaystyle [2\pi \dfrac{1}{54}(1 + 9(3)^{4})^{3/2}] - [2\pi \dfrac{1}{54}(1 + 9(2)^{4})^{3/2}]$