# Thread: Area of surfaces of revolution

1. ## Area of surfaces of revolution

HELP! I'm in Calc BC, i need step by step help

Problem: Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

41. y=((x^3)/6)+((1)/(2x)), [1,2]

Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the y-axis

43. y=(x^1/3)+2

x values [2,8]

y values [3,4]

2. Ok start with the formula for the SA of a solid of revolution around the x-axis. $\displaystyle S.A. = \int_{a}^{b} 2\pi f(x)\sqrt{1+f'[x]^2}dx$ Can you setup the integrals and try them?

3. i know the eqn and all that...i just need help with the step by step...

4. I'm on this to!

I did it like this (i'd get a second opinion since i'm new to this topic):

5. Originally Posted by Jameson
Ok start with the formula for the SA of a solid of revolution around the x-axis. $\displaystyle S.A. = \int_{a}^{b} 2\pi{\color{red}f[x]}\sqrt{1+f'[x]^2}dx$ Can you setup the integrals and try them?
You forgot that vital term!

--Chris

6. Originally Posted by vstexas09
HELP! I'm in Calc BC, i need step by step help

Problem: Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

41. y=((x^3)/6)+((1)/(2x)), [1,2]

Let $\displaystyle y=f(x)=\tfrac{1}{6}x^3+\frac{1}{2x}$

Thus, $\displaystyle f'(x)=\tfrac{1}{2}x^2-\frac{1}{2x^2}$

Therefore, $\displaystyle [f'(x)]^2=\left(\tfrac{1}{2}x^2-\tfrac{1}{2}x^{-2}\right)^2=\tfrac{1}{4}x^4-\tfrac{1}{2}+\tfrac{1}{4}x^{-4}$

Plugging this into the surface area equation, we get:

\displaystyle \begin{aligned}S.A.&=2\pi\int_1^2 f(x)\sqrt{1+\left(\tfrac{1}{4}x^4-\tfrac{1}{2}+\tfrac{1}{4}x^{-4}\right)}\,dx\\ &=2\pi\int_1^2 f(x)\sqrt{\tfrac{1}{4}x^4+\tfrac{1}{2}+\tfrac{1}{4 }x^{-4}}\,dx\end{aligned}

Note that $\displaystyle \tfrac{1}{4}x^4+\tfrac{1}{2}+\tfrac{1}{4}x^{-4}$ is a perfect square: $\displaystyle \tfrac{1}{4}x^4+\tfrac{1}{2}+\tfrac{1}{4}x^{-4}=\left(\tfrac{1}{2}x^2+\tfrac{1}{2}x^{-2}\right)^2$

Thus, the integral becomes:

\displaystyle \begin{aligned}S.A.&=2\pi\int_1^2 f(x)\sqrt{\left(\tfrac{1}{2}x^2+\tfrac{1}{2}x^{-2}\right)^2}\,dx\\ &=2\pi\int_1^2 f(x) \left(\tfrac{1}{2}x^2+\tfrac{1}{2}x^{-2}\right)\,dx\end{aligned}

Since, $\displaystyle f(x)=\tfrac{1}{6}x^3+\frac{1}{2x}$, we see that the integral transforms into

\displaystyle \begin{aligned}S.A.&=2\pi\int_1^2\left(\tfrac{1}{1 2}x^5+\tfrac{1}{3}x^3+\tfrac{1}{4}x^{-3}\right)\,dx\\ &=\frac{\pi}{6}\int_1^2 \left(x^5+4x^2+3x^{-3}\right)\,dx\end{aligned}

Take it from here. You should be able to do this!

I hope this helps!

--Chris

7. Originally Posted by Chris L T521
You forgot that vital term!

--Chris
You are right! That was a very bad typo. Thank you. Fixed now.

8. darn it! That means i've just found the length of it!

I worked it out again and got 115pi/36.

I would post my working except I did it on a whiteboard.