Arc length of (sqrt(y))/3)*(y-3) 1

Question

Arc length of (sqrt(y))/3)*(y-3) 1<y<9

experimentx · Answer

$$ f(y) = {\sqrt{y} \over 3(y - 3)}, 1<x<2  \implies f(x) = {\sqrt x \over 3(y - 3)}, 1<x<2$$
$$ \text{Arc Length}= \int_1^2\sqrt{ 1 + {f'(x)^2}}dx $$

anonymous · Answer

Yeah I know that, but I got up to f(y) part then I just don't know how to break it up when I square it to ingrate.

experimentx · Answer

http://www.wolframalpha.com/input/?i=Int [Sqrt[1+%2B+%28D[Sqrt[x]%2F%283%28x-3%29%29%2C+x]%29^2]%2C+{x%2C+1%2C+2}]

anonymous · Answer

I don't see anything, but I want to know how to do it. step by step.

experimentx · Answer

I don't think you can do it :( http://www.wolframalpha.com/input/?i=Int%5BSqrt%5B1+%2B+%28D%5BSqrt%5Bx%5D%2F%283%28x-3%29%29%2C+x%5D%29%5E2%5D%2C+x%5D

anonymous · Answer

Well I got f'(y)=(1/2)sqrt y-(1/2sqrt y)

experimentx · Answer

i think the best approach to do this is numerical approach ...

anonymous · Answer

What do you mean?

experimentx · Answer

http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx the method is correct ... using that method, you can find the function for integration ... but the integration is not closed (according to wolfram) ...

anonymous · Answer

I wrote it wrong, it should be$$1/3\sqrt {y}(y-3)$$

experimentx · Answer

well same result http://www.wolframalpha.com/input/?i=Int%5BSqrt%5B1+%2B+%28D%5B1%2F%28Sqrt%5Bx%5D3%28x-3%29%29%2C+x%5D%29%5E2%5D%2C+x%5D

anonymous · Answer

No, I got something.

anonymous · Answer

http://www.wolframalpha.com/input/?i=Int%5BSqrt%5B1+%2B+%28D%5BSqrt%5Bx%5D%2F3%28x-3%29%29%2C+x%5D%29%5E2%5D%2C+x%5D

experimentx · Answer

you got the wrong function
$$ f(x) = {\sqrt{x}(x-3) \over 3}$$

anonymous · Answer

That's not it. It's what I put in.

anonymous · Answer

(sqrt(y))/3)*(y-3)

experimentx · Answer

the best approach to solve this problem is to solve it numerically

anonymous · Answer

What do you mean by that?

experimentx · Answer

integration of this function http://www.wolframalpha.com/input/?i=Sqrt%5B1+%2B+%28D%5BSqrt%5Bx%5D%2F%283%28x-3%29%29%2C+x%5D%29%5E2%5D is a non elementary integral which you cannot expressed in terms of algebraic terms

experimentx · Answer

check this out http://www.wolframalpha.com/input/?i=integrate+sqrt%281%2B%281%2F%286+%28-3%2Bx%29+sqrt%28x%29%29-sqrt%28x%29%2F%283+%28-3%2Bx%29%5E2%29%29%5E2%29+

experimentx · Answer

but since it is a definite integral .. you can solve it numerically using Riemann sum or Trapezoidal rule or Simpson's method or using computer http://www.wolframalpha.com/input/?i=integrate+sqrt%281%2B%281%2F%286+%28-3%2Bx%29+sqrt%28x%29%29-sqrt%28x%29%2F%283+%28-3%2Bx%29%5E2%29%29%5E2%29++from+1+to+2

anonymous · Answer

No, that's not it. It's onethird squareroot of y times (y minus 3)

experimentx · Answer

$$ f(y) = {\sqrt{y} (y - 3) \over 3}$$

anonymous · Answer

no

experimentx · Answer

there isn't solution for 
$$ \int \sqrt{ 1 + \left ( {d \over dy } \left(\sqrt{y} \over 3 (y -3)\right )\right )^2} dy$$

anonymous · Answer

That's not it. (Sqrt[x]/3)(x-3)

anonymous · Answer

$$(\sqrt{y}/3)(y-3)$$

experimentx · Answer

what is that??
isn't it|dw:1342327003587:dw|