Length of Polar Curve 5(theta)^2 for 0 <= (theta) <= sqrt[21]

[dbfn2]

Hi,

I am stuck on an example problem and have no idea where they are getting the value 5/2 from. If someone could please let me know I would appreciate it as I understand the rest of the problem.

Find the length of the following polar curve. The spiral \(\displaystyle {5\theta}^2\) for \(\displaystyle 0 \leq \theta \leq \sqrt{21}\)

They proceed to work the problem out and I am able to follow up until they change the limits of integration and suddenly the 5 out front is divided by 2. Why is this happening?

\(\displaystyle L = \int_{0}^{\sqrt{21}}5\theta\sqrt{\theta^2 + 4} d\theta\)

\(\displaystyle = \int_{4}^{25}\frac{5}{2}\sqrt{v} dv\)

After this step they just integrate and I am able to follow along and get the answer of 195.

 

[HallsofIvy]

In addition to the fact that "suddenly the 5 out front is divided by 2", you should notice that the \(\displaystyle \theta^2+ 2\) inside the radical suddenly becomes "v". That should tell you immediately that they used the substitution \(\displaystyle v= \theta^2+ 4\). Then \(\displaystyle dv= 2\theta d\theta\) so that \(\displaystyle \theta d\theta= \frac{1}{2}dv\).

With that substitution, \(\displaystyle \int 5\sqrt{\theta^2+ 24}\theta d\theta\) becomes \(\displaystyle \int 5\sqrt{v}\frac{1}{2}dv= \frac{5}{2}\int\sqrt{v}dv\).

As for the limits of integration, when \(\displaystyle \theta= 0\), \(\displaystyle v= 0^2+ 4= 4\) and when \(\displaystyle \theta= \sqrt{21}\), \(\displaystyle v= (\sqrt{21})^2+ 4= 25\).

 

[dbfn2]

In addition to the fact that "suddenly the 5 out front is divided by 2", you should notice that the \(\displaystyle \theta^2+ 2\) inside the radical suddenly becomes "v". That should tell you immediately that they used the substitution \(\displaystyle v= \theta^2+ 4\). Then \(\displaystyle dv= 2\theta d\theta\) so that \(\displaystyle \theta d\theta= \frac{1}{2}dv\).

With that substitution, \(\displaystyle \int 5\sqrt{\theta^2+ 24}\theta d\theta\) becomes \(\displaystyle \int 5\sqrt{v}\frac{1}{2}dv= \frac{5}{2}\int\sqrt{v}dv\).

As for the limits of integration, when \(\displaystyle \theta= 0\), \(\displaystyle v= 0^2+ 4= 4\) and when \(\displaystyle \theta= \sqrt{21}\), \(\displaystyle v= (\sqrt{21})^2+ 4= 25\).

The dv thing is what I originally thought but dismissed it when working with a similar problem since I couldn't reproduce the answer given.


\(\displaystyle r = 2{\theta}^2\) for \(\displaystyle 0\leq\theta\leq\sqrt{21}\)

I get to the same place as the above:

\(\displaystyle = \frac{4}{2}\int_{4}^{25}\sqrt{v} dv\)

But when I evaluate I end up with:

\(\displaystyle \frac{4}{2}[ \frac{250}{3} - \frac{16}{3}]\)

\(\displaystyle 2\cdot(78) = 156\)

The correct answer is 78 however I am forced to double it with the outside value of 2. I don't see why this problem is any different from the previous one which I was able to solve for after integration. I know it seems silly to ask these things but I do not see where I am messing up.

 

[dbfn2]

Actually on second thought it doesn't matter - for whatever reason it was just that one problem I couldn't figure out even though it's the exact same process. Maybe it's a mistake in the reading material. Thanks for the help HallsofIvy.