Area under a curve, difference between dy and dx I am trying to find the area of $ y = 1 $ and $y = x^\frac{1}{4}$ from 0 to 1 and revolving around $ x = 1$
In class we did the problem with respect to y, so from understanding that is taking the "rectangles" from f(y) or the y axis. I was wondering why not just do it with respect to x, it would either be a verticle or horizontal slice of the function but the result would be the same. I was not able to get the correct answer for the problem but I am not sure why.
Also one other question I had about this, is there a hole at 1,1 in the shape? The area being subtracted is defined there so shouldn't that be a hole since we are taking that area away? Both function at 1 are 1.
 A: I expect you have drawn a picture, and that it is the region below $y=1$, above $y=x^{1/4}$, from $x=0$ to $x=1$ that is being rotated about $x=1$. When you rotate, you get a cylinder with a  kind of an upside down bowl carved out of it, very thin in the middle.  You have asked similar questions before, so I will be brief.
It is probably easiest to do it by slicing parallel to the $x$-axis. So take a slice of thickness $dy$, at height $y$. We need to find the area of cross-section. 
Look at the cross-section. It is a circle with a circle removed. The outer radius is $1$, and the inner radius is $1-x$. So the area of cross-section is $\pi(1^2-(1-x)^2)$. We need to express this in terms of $y$. Note that $x=y^4$. so our volume is
$$\int_0^1 \pi\left(1^2-(1-y^4)^2\right)\,dy.$$
I would find it more natural to find the volume of the hollow part, and subtract from the volume of the cylinder.
You could also use shells. Take a thin vertical slice, with base going from $x$ to $x+dx$, and rotate it. At $x$, we are going from $x^{1/4}$ to $1$. The radius of the shell is $1-x$, and therefore the volume is given by
$$\int_0^1 2\pi(1-x)(1-x^{1/4})\,dx.$$
Multiply through, and integrate term by term.  Not too bad, but slicing was easier.
