I need to find the volume of revolution over axis $\ x+1 = 0 \ $ of the area given by $\ y = \frac{1}{1+x}$ and $y= 0, \ x = 1, \ x = 0$
What I've been told by my professors is to move the axis $\ x = -1$ to $\ x = 0$ but I also have to move the function accordingly, and that's the thing, I don't really get if it's actually +1 or -1, because I thought that if I'm moving to the right, I had to add 1.
So what I've done is $f(x) = \frac{1}{1+x}$ if axis is $\ x = -1,$ so $\ f^{*}(x) = f(x + 1)$ if axis is $\ x = 0$ but I believe this way is actually wrong.
And then use $V =2π \int_{1}^{2} x\ f^{*}(x) \ \ dx$.
I'd appreciate it a lot if someone could illustrate this so that it was easier to understand.
Thanks in advance.
HINT
By shell method we have that
$$V=2\pi\int_0^1 (x+1)\cdot\frac1{x+1}dx$$
As an alternative we can make a translation of rotation axis x=-1 to the origin that is
then the problem becomes to find the volume of revolution over axis $u = 0 \ $ of the area given by $\ v = \frac{1}{u}$ and $v= 0, \ u = 2, \ u = 1$ that is
$$V=2\pi\int_1^2 u\cdot\frac1{u}du$$