Finding an arc length, and why isn't latex working for me?

[Stevecgz]

Finding an arc length

I am attempting to find the arc length of y = cuberoot[x] between (1,1) and (8,2).

I solved the integral from 1 to 2 of sqrt[1+(3y^2)^2]dy. I used a formula from a table of integrals in my text to solve this integral. The solution I get is 68.19. I can see that this is not a reasonable answer. Is my setup incorrect or am I solving the integral incorrectly?

Thanks for any help.

Steve

 

[iNCREDiBLE]

Error vv

 

[Gokul43201]

The set-up is okay. Looks like you're solving the integral incorrectly. Perhaps, if you show the working, someone will spot the error.

 

[Fermat]

Why isn't latex working for me

I thought it wasn't working for me either. But it seems to be not working properly only in the preview pane, when you are composing your thread/response.

When posted, the latex code works properly (at least for me it did

You can hit the edit button on your post and edit it there as need be.

 

[Stevecgz]

You're right Fermat, I was only trying it in the preview page, but it's working now.

Thanks Gokul, this is how I got to my answer.

This is the formula I used from my text:

$$\int \sqrt {a^2 + u^2} du = \frac u 2 \sqrt {a^2 + u^2} + \frac {a^2} {2} \ln{(u + \sqrt {a^2 + u^2})} + C$$

I used a = 1 and u = 3y^2, and that is how I came up with 68.19.

I'm trying to find a method to solve this integral without using that formula but I am having trouble. I can't see any logical u substitution that would work. I have tried a trigonometric substitution, with $$3y^2 = \tan{\theta}$$, but when I replace $$dy$$ with $$d\theta$$ I'm left with a harder integral than I started with. Thanks again for any help.

Steve

 

[Fermat]

have you tried,

u = a.sinht ?

 

[Fermat]

I think the integal you derived

$$\int \sqrt {a^2 + u^2}\ du$$

is wrong.

If the integral is,

$$\int \sqrt {1 + (3y^2)^2}\ dy$$

then the substitution u = 3y² gives,

$$\int \sqrt {1 + u^2}\ dy$$

but

du = 6y dy

or

dy = du/6y = √3.du/(6√u)

which givves the integral as,

$$\int \sqrt {1 + u^2}\ \sqrt{3}du/(6\sqrt{u})$$

Also, I put your origianl integral,

$$\int \sqrt {1 + (3y^2)^2}\ dy$$

into this http://integrals.wolfram.com/" and got back exotic formulas (I think they're called)

I don't know how to evaluate those, sorry

 

[Stevecgz]

Also, I put your origianl integral,
$$\int \sqrt {1 + (3y^2)^2}\ dy$$
into this http://integrals.wolfram.com/" and got back exotic formulas (I think they're called)
I don't know how to evaluate those, sorry

I tried it at a similar site and got back some crazy stuff that didn't mean much to me.

I'm still not getting anywhere with this, I'm really curious about what I'm doing wrong. I'm sure it's just a simple mistake.

I guess I'll start working on another method and see if I get anywhere trying to find:

$$\int_{1}^{8} \sqrt{1 + \left[\frac {1}{3\sqrt[3]{x^2}}\right]^2} dx$$

Steve

 

[Stevecgz]

Is this integral not elementary? Do I need to use approximate integration?

Steve

 

[Stevecgz]

btt...