1. ## Surface Area Problem

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

$y = x^{3}$

$0 \leq x \leq 3$

$y' = 3x^{2}$

$S = \int_{0}^{3} 2\pi (x^{3}) \sqrt{1 + (3x^{2})^{2}} dx$

$S = 2\pi \int_{0}^{3} (x^{3}) \sqrt{1 + 9x^{4}} dx$

$S = 2\pi \int_{0}^{3}(x^{3}) \sqrt{1 + 9x^{4}} dx$

Use trig stuff here?

2. ## Re: Surface Area Problem

You also need to convert the \displaystyle \begin{align*} x^3 \end{align*} from your trigonometric substitution.

3. ## Re: Surface Area Problem

Originally Posted by Prove It
You also need to convert the \displaystyle \begin{align*} x^3 \end{align*} from your trigonometric substitution.
What do you mean about the $x^{3}$ and how would you go about trig sub. considering we have $1 + 9x^{4}$ This going to the 4th power not 2nd as required to use $a\tan\theta$

4. ## Re: Surface Area Problem

Since you have edited your original post I can't really state what we originally had. Anyway, you have correctly gotten to \displaystyle \begin{align*} S = \int_0^3{ 2\pi\left( x^3 \right) \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x} \end{align*}. From here I would do...

\displaystyle \begin{align*} \int_0^3{ 2\pi \left( x^3 \right) \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x } &= 2\pi \int_0^3{ x^3 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x } \\ &= \frac{\pi}{3} \int_0^3{ x^2 \, \sqrt{1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } \\ &= \frac{\pi}{9} \int_0^3{ 3x^2 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } \end{align*}

Now I would substitute \displaystyle \begin{align*} 3x^2 = \tan{ \left( \theta \right) } \implies 6x\,\mathrm{d}x = \sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}, and also note that when \displaystyle \begin{align*} x = 0, \theta = 0 \end{align*} and when \displaystyle \begin{align*} x = 3, \theta = \arctan{ \left( 27 \right) } \end{align*}, the integral becomes

\displaystyle \begin{align*} \frac{\pi}{9} \int_0^3{ 3x^2 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } &= \frac{\pi}{9} \int_0^{\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*}

Can you go from here?

5. ## Re: Surface Area Problem

Originally Posted by Jason76
Use trig stuff here?
You don't need a trigonometric substitution.
Use the substitution $\sqrt{1+9x^4}=t\Rightarrow 1+9x^4=t^2\Rightarrow 36x^3dx=2tdt\Rightarrow x^3dx=\frac{1}{18}tdt$.
You can continue from here.

6. ## Re: Surface Area Problem

Originally Posted by red_dog
You don't need a trigonometric substitution.
Use the substitution $\sqrt{1+9x^4}=t\Rightarrow 1+9x^4=t^2\Rightarrow 36x^3dx=2tdt\Rightarrow x^3dx=\frac{1}{18}tdt$.
You can continue from here.
Yes you DO need a trigonometric substitution here. When you resubstitute in for the x^3 term, you will end up with another \displaystyle \begin{align*} \sqrt{ a^2 + t^2} \end{align*} form.

7. ## Re: Surface Area Problem

This problem needs u substitution, not trig sub. It's just like an arc length problem before. Same numbers.

$y = x^{3}$

$0 \leq x \leq 3$

$y' = 3x^{2}$

$S = \int_{0}^{3} 2\pi (x^{3}) \sqrt{1 + (3x^{2})^{2}} dx$

$S = 2\pi \int_{0}^{3} (x^{3}) \sqrt{1 + 9x^{4}} dx$

$S = 2\pi \int_{0}^{3}(x^{3}) \sqrt{1 + 9x^{4}} dx$

$S = 2\pi \int_{0}^{3}(x^{3}) (1 + 9x^{4})^{1/2} dx$

$S = 2\pi \int_{0}^{3}(x^{3}) (u)^{1/2} dx$

$u = 1 + 9x^{4}$

$du = 36x^{3} dx$

$\dfrac{1}{36} = x^{3} dx$

$= 2\pi \dfrac{1}{54}(u)^{3/2}$ evaluated at upper bound 3 and lower bound 0

$= 2\pi \dfrac{1}{54}(1 + 9x^{4})^{3/2}$ evaluated at upper bound 3 and lower bound 0

$[2\pi \dfrac{1}{54}(1 + 93)^{4})^{3/2}] - [2\pi \dfrac{1}{54}(1 + 9(0)^{4})^{3/2}]$