I have to find the volume of the bounded region by the following functions:
$$y=\frac{1}{x^3}$$ $$y=0$$ $$x=1$$ $$x=2$$ $$\mathbf{Revolving\ around\ the \ axis:}$$ $$x=-1$$
The method that one has to use is cylindrical shells, but if you could do it both methods slicing, and cylindrical.
I have this integral for cylindrical shells is:
$$V=2\pi \int_\limits{1}^{2}((x+1)(\frac{1}{x^3}))dx$$
But I did it using slicing as well which lead me to this integral:
$$V=\pi \int_\limits{\frac{1}{8}}^{1}((\sqrt[3]{\frac{1}{y}}+1)^2-1^2)dy$$
Would it be possible, being that I know they should be equal the displacing of the axis of revolution is being quite trouble some, could you go in detail to how the axis is impacting both?