MHB Is my integral for the volume of a solid by rotating y=2/x around y=-1 correct?

[stripedcat]

Okay, so, we have

y=2/x, y=0, x=1, x=3 and it's rotating around y=-1

As I understand it... This means

(2/x +1)^2-(0+1)^2
(4/x^2 + 4/x)

The integral of that is

-4/x + 4ln(x)

Am I right so far?

Plug in the 3, and then the 1

-4/3 + 4ln(3) - -4+4ln(1)

Then the result of that is multiplied by pi

This is where I'm missing something. I tried wolfram, it's coming up with pi(8/3+log(3))?

Is that right? My first answer was pi(8/3+log(81)) but that wasn't right.

 

[MarkFL]

You have the right answer, if you just simplify some...combine the two numbers and observe that ln(1) = 0. I'm not sure what command you used with W|A, but that result is not correct. To avoid confusion, you should write:

-4/3 + 4ln(3) -( -4+4ln(1))

 

[stripedcat]

It does simply down to 8/3+4ln(3)

But what does that become when multiplied by pi?

One of several options wolfram shows is indeed pi(8/3+log(81))

It also comes back with 8pi/3 + 4pi log(3)

And 4/3pi(2+3log(3))

I'm trying to verify I at least got up to a certain point right.

 

[MarkFL]

Note that:

$$4\ln(3)=\ln\left(3^4\right)=\ln(81)$$

They are all just different ways to write the same thing.