consider the following question:
find the volume of a solid bounded by $y=5/2-x$ and $y=1/x$, around $x=-1$.
Know the answer is:
$$\int_{0.5}^{2} \left(\frac{5}{2}-y-1\right)^2 -\left(\frac{1}{y}-(-1)\right)^{2} dy$$
I am wondering if I write the integral with respect of $x$, then my answer is correct?
i.e:
$$\int_{0.5}^{2} \left(\frac{5}{2}-x-1\right)^2 -\left(\frac{1}{x}-(-1)\right)^{2} dx$$
I get the same answer. but is it geometrically correct?
Once you have an integral, you can do a change of variables without changing the value. However, for setting up the integral, there is a geometric picture, and the variables have meaning. When you use the method of washers, you are cutting a solid of revolution into slices perpendicular to the axis of rotation, integrating the cross sectional area, and so you must view the cross sectional area as a function of the variable that is changing along the axis of rotation, and you must make an integral with respect to that variable.
When you get to finding volumes using the method of shells, if you have a problem that can be attacked with either method, one method will involve an integral with respect to x and the other with respect to y. It will be terribly confusing if you don't keep your variables straight.