# Thread: Arc Length and Area Surface by revolution

1. ## Arc Length and Area Surface by revolution

y=1/6(x³) + 1/(2x) , 1 ≤ x ≤ 2

I don't really understand how to solve this...

2. arc length ...

$\displaystyle S = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

surface area ...

$\displaystyle A = \int_a^b r(x) \cdot \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$
where $\displaystyle r(x)$ is the radius of revolution as a function of $\displaystyle x$

3. ok for the arc length i have

∫√[1- [(x⁴-1)/(2x²)]²] dx from 1 to 2

∫√[1- [(x⁸-2x⁴+2)/(4x⁴)] dx from 1 to 2

i can't get past that.

For the srface area i have

2∏∫ [(x³)/6 + 1/(2x)] √[1- [(x⁸-2x⁴+2)/(4x⁴)] dx

and I can't get past that either.

4. Originally Posted by hlpplz
y=1/6(x³) + 1/(2x) , 1 ≤ x ≤ 2
$\displaystyle \frac{dy}{dx} = \frac{x^2}{2} - \frac{1}{2x^2}$

$\displaystyle \left(\frac{dy}{dx}\right)^2 = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}$

$\displaystyle 1 + \left(\frac{dy}{dx}\right)^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2$

$\displaystyle \sqrt{\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2} = \frac{x^2}{2} + \frac{1}{2x^2}$

$\displaystyle S = \int_1^2 \frac{x^2}{2} + \frac{1}{2x^2} \, dx$

$\displaystyle A = \pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)^2\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx$

can you finish ?

5. how did you get

$\displaystyle A = \pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)^2\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx$

i thought the equation for surface area was

2∏∫ f(x) √[1+f'(x)] dx

wouldn't that make it be

2∏$\displaystyle \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx$

then,

2∏$\displaystyle \int_1^2 \frac{1}{2}\left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx$

thus,

∏$\displaystyle \int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx$

6. you're right ... I had area of a circle on the brain instead of circumference.

corrected.

$\displaystyle A = 2\pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx$

7. so the i have

∏$\displaystyle \int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx$

∏$\displaystyle \int_1^2 \left(\frac{x^5}{3} + \frac{x^3}{3x^2} + \frac{x^2}{x} + \frac{x}{x^3}\right) \, dx$

∏$\displaystyle \int_1^2 \left(\frac{x^5}{3} + \frac{x}{3} + {x} + \frac{1}{x^3}\right) \, dx$

∏$\displaystyle \int_1^2 \left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx$

∏$\displaystyle \left(\int_1^2 \frac{x^5}{3} + \int_1^2 \frac{4x}{3} + \int_1^2 \frac{1}{x^3} \, dx \right)$

am i right so far?

8. you can only multiply one of the two factors in the integrand by 2, not both.

$\displaystyle \pi \int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx$

9. ok from this

$\displaystyle \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx$

what i did was just multiply it all out and got

2∏$\displaystyle \int_1^2 \left(\frac{x^5}{12} + \frac{x^3}{12x^2} + \frac{x^2}{4x} + \frac{1}{4x^3}\right) \, dx$

2∏$\displaystyle \int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{x}{4} + \frac{1}{4x^3}\right) \, dx$

2∏$\displaystyle \int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{3x}{12} + \frac{1}{4x^3}\right) \, dx$

2∏$\displaystyle \int_1^2 \left(\frac{x^5}{12} + \frac{4x}{12} + \frac{1}{4x^3}\right) \, dx$

2∏$\displaystyle \int_1^2 \frac{1}{4}\left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx$

∏/2$\displaystyle \int_1^2 \frac{1}{4}\left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx$

is that better?

10. $\displaystyle 2\pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx$

$\displaystyle 2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x^3}{12x^2} + \frac{x^2}{4x} + \frac{1}{4x^3}\right) \, dx$

$\displaystyle 2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{x}{4} + \frac{1}{4x^3}\right) \, dx$

$\displaystyle 2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x}{3} + \frac{1}{4x^3}\right) \, dx$

integrate at this point. don't mess with the algebra anymore ... too big a chance of making a bonehead error.

11. ok so for the final result i was able to get

$\displaystyle \frac{41\pi}{16}$

is that correct?

12. Originally Posted by hlpplz
ok so for the final result i was able to get

$\displaystyle \frac{41\pi}{16}$

is that correct?
my calculator says $\displaystyle \frac{47\pi}{16}$

13. and for the arc length i ended up getting

$\displaystyle \frac{17}{12}$

is that right?

14. you're done.