Calc III Problem Integrating Sq Rt's in Arc Length Formula

[yUNeeC]

Hey guys, I'm studying for a test in calc 3 tomorrow and have run into a problem. On the practice test we have a problem "Find the length of the curve: r=theta^2, 0≤theta≤pi/2"

I know the length of a curve in polar coordinates is int(sqrt(r^2 + (dr/dtheta)^2))dtheta...but when I get to where I have to integrate sqrt(theta^4 + 4theta^2) I become very stuck. What is the procedure for handling this square root integration?

Thanks for any help.

 

[Svalbard]

Hi,

so you are trying to integrate:

\int\sqrt{t^4 + 4t^2} dt = \int t\sqrt{t^2 + 4}dt

Substitute t^2+4 = u

=> dt = du/2t

t cancels out:
\int t\sqrt{t^2 + 4}dt = \frac{1}{2} \int \sqrt{u} du = \frac{u^{\frac{3}{2}}}{3} + c

\Rightarrow \int\sqrt{t^4 + 4t^2} dt = \frac{(t^2+4)^{\frac{3}{2}}}{3} + c

is this what you are asking for?

 

[yUNeeC]

Wow I feel stupid. Haha thanks a bunch.