Question

The region bounded by y=1/(x^2+2x+5),y=0, x=0, and x=1, is revolved about the y-axis. Find the volume of the resulting solid.

Answer 1

Using the shell method, the integral set up is
\[V=2\pi\int_0^1\frac{1}{x^2+2x+5}\;dx\]
When evaluating the integral, I suggest completing the square in the denominator.

Answer 2

thank you very much

Answer 3

how would i solve for C though?

Answer 4

You mean the constant of integration? That isn't an issue when evaluating `Definite Integrals`. No C to worry about.

If you wanted to put a +C just to see what would happen ~ You'd notice that when you evaluate your function at the upper and lower limits, the C's will just subtract out.

Answer 5

so I got (1/2)arctan((x+1)/2) when I took the integral

Answer 6

would and I tried pluging in the y(1)-y(0) and it says my answer is wrong

Answer 7

what am I doing wrong?

Answer 8

mm one sec, lemme work out the derivative, maybe just off by a fraction.

Answer 9

Yah I'm coming up with,\[\large \color{royalblue}{\frac{1}{4}}\arctan\left(\frac{1}{2}(x+1)\right)\]
Maybe check your work, I'll check mine too. I kinda rushed through it lol.

Answer 10

ok, will do

Answer 11

I keep getting the same answer

Answer 12

Hmm ok let's see if we can figure out where the mistake is.
Oh wait.. I forgot about the 2pi in front -_- ugh I'm dumb...

I don't see a pi in yours after integrating, did you miss that also? :o

Answer 13

just didnt write it in...I have 2pi (1/2)arctan((x+1)/2)

Answer 14

then i did 2pi(y(1)-y(0))

Answer 15

\[\large 2\pi\int\limits_0^1\frac{1}{x^2+2x+5}\;dx \qquad = \qquad 2\pi\int\limits_0^1\frac{1}{(x+1)^2+4}\;dx\]So you were able to get this correct? Then we make the following substitution,\[\large x+1=2\tan \theta\]\[\large dx=2\sec^2\theta d \theta\]
There's actually a formula you can memorize for this, but I find it helpful to understand the actual steps, in case you run into tougher problems that are not as straight forward.

Plugging everything in gives us,\[\large 2\pi\int\limits\limits_0^1\frac{1}{4\tan^2 \theta+4}\;\left(2\sec^2\theta d \theta\right)\]Which simplifies down to,\[\large \pi\int\limits_0^1 d \theta\]

Answer 16

Blah ignore those limits* We plan on switching back to x anyway.

Answer 17

yes and I still get arctan ((x+1)/2)/2

Answer 18

so to where are the limits applied to? are the limits 1 and 0?

Answer 19

Hmm where is that outer 1/2 coming from?
Integrating will give us,\[\large \pi \theta\]Putting this back in terms of x will give us,\[\large \pi \arctan\left(\frac{1}{2}(x+1)\right)\]

Yah it looks like our limits for x are 0 and 1. Not theta.

Answer 20

ok I tried putting the limits and it says that it is still wrong

Answer 21

so i did pi arctan( (1+1)/2) -pi arctan( (0+1)/2)

Answer 22

where am i going about it wrong?

Answer 23

I checked it on wolfram real quick to make sure we weren't way off.
That in fact `is` correct.

For the first term, recall the arctan(1) gives us a special angle, namely, \(\dfrac{\pi}{4}\).\[\large \pi \arctan\left(1\right)-\pi \arctan\left(\frac{1}{2}\right)\]\[\large \pi\left[\frac{\pi}{4}-\arctan\left(\frac{1}{2}\right)\right]\]

Which is equivalent to what Wolfram has.
http://www.wolframalpha.com/input/?i=integral+from+0+to+1+2pi+1%2F%28x%5E2%2B2x%2B5%29+dx

Answer 24

If that's not correct, then maybe one of two things is going on.
~Maybe the problem is telling you to round your answer to a decimal.
~Or maybe there's a mistake in the shells setup at the start.
I didn't bother to think about that, it just looked correct XD lol

Answer 25

darn...says that is incorrect too. Do i need to maybe multiply by the volume of a specific solid?

Answer 26

Oh no!! This isn't good! I think there was a mistake in the shells setup! :(Our region looks something like this, if we spin a slice around the y-axis we get,