Integration: arclength for x = (1/3)sqrt[y](y - 3), 1<=y<=9

[scrum]

I've got this arclength problem

And I did the first bits with the arclength formula (

)


until I reach a bit where I have Integration from 9 to 1 of Sqrt(.25y+.25y^-1 -.5)

I don't know how to integrate that. U substitution fails because .25y^-1 turns into -.25y^-2 in the Du.

I do know from using a computer that I'm on the right track as the area under that curve from 9 to 1 was the correct answer, and that the answer is 10.66667

 

[galactus]

Re: Integration

If you square the derivative you get \(\displaystyle \frac{(y-1)^{2}}{4y}\)

Add 1 and take the square root and get:

\(\displaystyle \frac{\sqrt{\frac{1}{y}}(y+1)}{2}\)

So, you must integrate \(\displaystyle \frac{1}{2}\int_{1}^{9}\frac{y+1}{\sqrt{y}}dy\)

Now, let \(\displaystyle u=\sqrt{y}, \;\ 2du=\frac{1}{\sqrt{y}}dy, \;\ u^{2}=y\)

Make the subs and get:

\(\displaystyle \int_{1}^{3}(u^{2}+1)du\)

 

[soroban]

Re: Integration

Hello, scrum!

Check your algebra . . .

\(\displaystyle x \;=\;\frac{1}{3}\sqrt{y}(y-3)\quad 1 \leq y \leq 9\)

\(\displaystyle \text{We have: }\;x \;=\;\frac{1}{2}\left(y^{\frac{3}{2}} - 3y^{\frac{1}{2}}\right)\)

\(\displaystyle \text{Then: }\;\frac{dx}{dy} \;=\;\frac{1}{3}\left(\frac{3}{2}y^{\frac{1}{2}} - \frac{3}{2}y^{-\frac{1}{2}}\right) \;=\;\frac{1}{2}\left(y^{\frac{1}{2}} - y^{-\frac{1}{2}}\right)\)

. . . . \(\displaystyle \left(\frac{dx}{dy}\right)^2 \;=\;\frac{1}{4}\left(y^{\frac{1}{2}} - y ^{-\frac{1}{2}}\right)^2 \;=\;\frac{1}{4}\left(y - 2 + y^{-1}\right)\)

. . \(\displaystyle 1 + \left(\frac{dx}{dy}\right)^2 \;=\; 1 + \frac{1}{4}\left(y - 2 + y^{-1}\right) \;=\;\frac{1}{4}\left(y + 2 + y^{-1}\right) \;=\; \frac{1}{4}\left(y + y^{-1}\right)^2\)

\(\displaystyle \text{Therefore: }\;\sqrt{1 + \left(\frac{dx}{dy}\right)} \;=\;\frac{1}{2}\left(y + y^{-1}\right)\)


Now try it . . .