Figuring out the Radius when finding the volume of solid rotations

Question

When rotating a function around the x-axis, the radius of thin slice will be the same as the function itself. But, when rotating around the y-axis, you need to solve for x to figure out what that radius is. Why?
This graph was taken from Khan Academy. If I rotate R around the Y-axis, the radius of a thin horizontal slice will be the distance from the function to the y-axis. Why isn't that simply the function itself? How is it really $x=\ln (y)$?

Answer 1

If you have a point $(x,y)$ on the curve and rotate the curve around the $X$ axis the distance from the center of rotation is $y$, which is $f(x)$. If you rotate the curve around the $Y$ axis, the distance from the center of rotation is $x$. The horizontal lines at $y=2$ and $y=5$ show this. The points where the lines hit the curve are $(\log (2),2)$ and $(\log(5),5)$. As you are going to integrate with respect to $y$, you need the $x$ value at a given $y$, which is $x=f^{-1}(y)$. When $y=e^x, x=\log(y)$

Answer 2

A point on the curve is $P=(x, y)=(x, e^x)$. Its projection onto the $x$ axis is the point $A=(x, 0)$. Its projection onto the $y$ axis is $B=(0, e^x)$. You see thas $d(P, A)=e^x$ and $d(P,B) = x = \ln y$.