Finding Volume...How is the ArcTan[0] = 0?

pinkprincess08 · #1 $Old$ March 7th, 2009, 08:51 PM

I'm trying to find the volume of a tank that is created by rotating y=tan[x] around the y axis, 0<= y <= 6, using the disk method..
so I have:

pi * Integral from {0 to 6} of Arctan[y]^2.

We're supposed to use a Computer Algebra System(CAS) to solve it, but the CAS gives me an imaginary numbers in the answer and it's not making any sense. So I tried to solve by using the Fundamental Theorem of Calculus..but I get a completely different answer..

Also, how is ArcTan[0] = 0?
Tan[0] = sin[0]/cos[0] = 0 / 1 = 0 (i understand that..)
Arctan[0] = cos[0]/sin[0] = 1 / 0 ....(I dont understand how that's 0...am i doing anything wrong?)
Please help!

Mentia · #2 $Old$ March 7th, 2009, 10:27 PM

When I first encountered inverse trig functions I had the same problem. It turns out that

$arctan(x) \neq \frac{1 }{ tan(x)}$

but rather,

$arctan[tan(x)] = x$

So when you are taking the arctangent of a number, call it A, you are asking "The tangent of what number gives me A?"

So for arctan(0), you ask "The tangent of what number is zero?" And you note in your post that tan(0) = 0. Apparently the tangent of zero is itself zero. Then arctan(0) = 0.

As far as the volume goes, we have:

$y=tan(x) \Rightarrow x = arctan(y)$ ,

then the volume integral is:

$2 \pi \int_{0 }^{ 6} \int_{0 }^{arctan(y)} xdxdy =$

which is equal to just what you said in your post:

$\pi \int_{0 }^{6 } { arctan(y)}^{ 2} dy$

Now this is a horrible integral involving complex numbers, but in this case the imaginary part is zero. The best thing to do is just use the numerical integral solver on your calculator. You should get about 25.23 for the total volume.

pinkprincess08 · #3 $Old$ March 8th, 2009, 03:14 AM

Thanks alot. That helped solve the problem, as well as understand inverse trig functions.