# 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.

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It is easier to compute the volume with disks/washers than with cylindrical shells. Since you are revolving around $x=1$, if you try to set it up with respect to $x$ you will be using cylindrical shells; if you set it up with respect to $y$, you are using disks/washers. You can use either procedure to get the correct answer if you do them right, but one usually tries to use the method which is likely to be easier, not the one that is likely to be harder, if given the choice. – Arturo Magidin May 1 '12 at 3:51

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You are trying to find a volume generated by the strip as it rotates. Your limit will be over the thickness of the strip going to $0$. If the side which corresponds to this thickness is perpendicular to the axis of rotation, then the area is simply given by $\pi \ell^2\Delta$, where $\ell$ is the length/height, whichever is appropriate, and $\Delta$ the thickness; $\ell$ is often just the value of the function. (cont) – Arturo Magidin May 1 '12 at 4:00

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(cont) If the thickness is parallel to the axis of rotation, then you get a cylindrical shell, and the volume is the more complicated $2\pi r\ell\Delta$, where $r$ is distance to axis, $\ell$ is length/height of the slice (which ever is appropriate), and $\Delta$ the thickness. This usually involves computing two different quantities, $r$ and $\ell$, instead of a single one. Note you don't get a disk by rotating a vertical strip vertically: you get a hollow cylinder. Volumes of disks are easy; volumes of hollow cylinders are hard and not degree 1 on the thickness. – Arturo Magidin May 1 '12 at 4:03

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Like I said: you can use either one to compute volume, but when the strips are perpendicular to the axis of rotation, it usually leads to a simpler formula for your integration. If the strips are parallel to the axis of rotations, then when you try to compute the volume with an integral it usually leads to a harder formula. You cannot use the formula for disks with strips that are parallel to the axis of rotation, because those strips don't describe a disk when rotated, they describe a hollow cylinder. – Arturo Magidin May 1 '12 at 4:10

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Yes; imagine a pole in a merry-go-round. The shape it describes is neither a disk (a thick circle) nor a washer (a thick circle with a hole in the middle). It is like a hollow cylinder. So you don't compute the volume of that shape with the formula $\pi r^2\Delta$, because that's the volume of a disk, not the shape being described by the pole going around the merry-go-round. – Arturo Magidin May 1 '12 at 4:14
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## 1 Answer

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.

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