The example I was given is find the volume of the solid obtained by rotating the region enclosed by $y=\sqrt{64-x^2}$ and $y=6$ about the x-axis.
I first tried to integrate half of the region multiplied by 2 by writing the integral in terms of x, solving $2\int_{0}^{\sqrt{28}} 2\pi(\sqrt{64-x^2})(x)dx$
This gave me the wrong answer however. The right method was to write the integral in terms of y using $2\int_{6}^{8} 2\pi(\sqrt{64-y^2})(y)dy$
I see that they must be different because the bounds are different while the integrands are the same. But I can't understand visually how integrating the region left from right compared to bottom to top could give me a different answer. Does it have to do with $dx$ versus $dy$?
Could someone please tell me the reason why I must write an integral in terms of y when rotating a region about the x-axis?
Thanks for the help.
But you're not integrating a region. You're integrating a three-dimensional solid. The shell method is just a shortcut that lets you integrate it quicker.
It's not really left to right versus bottom to top. It's whether you integrate along the axis of rotation (using the disk or washer method) or whether you integrate from inside to outside in the direction away from the axis of rotation (using the shell method).
What you have done with your modification to the shell method for this problem is not like turning a cylinder on its side. In fact, it's wrong in so many ways that it's hard to describe what it is that you've actually done in a way that makes any geometric sense. The fact that this is not already obvious to you is evidence to me that you have not been taught anything about how the shell method actually works.
When you write the integral correctly, as
$$ \int_6^8 2 \pi y\left(\sqrt{64-y^2}\right)\,\mathrm dy $$
(I changed the order of multiplication in the integrand for purposes of exposition), the $\left(\sqrt{64-y^2}\right)\,\mathrm dy$ part can be interpreted as saying that you're going to make each "shell" in the shell method by rotating a thin rectangular strip of length $\sqrt{64-y^2}$ and width $\mathrm dy$ [note 1] around an axis to make a cylindrical shell of radius $y.$ In rotating this thin strip all the way around the axis to make the cylindrical shell, you move the strip a distance $2\pi y$ in a direction perpendicular to the surface of the strip, thereby creating a volume $2\pi y$ times the area of the strip [note 2], namely
$$ 2 \pi y\left(\sqrt{64-y^2}\right)\,\mathrm dy. $$
The integral then combines the volumes of the strips for values of $y$ from $6$ to $8$ [note 3].
But we have skipped an important step here: how did we decide what length the strip should have and what the radius of the shell should be?
So the very first thing we should do is to determine what region we are rotating around what axis. The graph below shows this region, which is a sector of a circle in the $x,y$ plane. We are supposed to rotate the gray region around the $x$ axis. The result will be a sphere with a cylindrical hole through the middle.
As an aid to visualization, this is what the solid of rotation should look like in the end:
You can recognize that the region in the $x,y$ plane (and hence also the sphere-with-hole) is symmetric across the $y$ axis and that you can compute the volume just as well by making a vertical cut through the region at the $y$ axis and only integrating the part on the right side of the cut, then doubling it.
Since we are rotating around the $x$ axis, the only way we are going to get a cylindrical shell is if the rectangle we are rotating has its long dimension horizontal. And for the shell to represent part of the volume of the solid object, the rectangle really should be inside the region that we rotate. Because you've already cut that region in half, we end up making a horizontal rectangle just in the half-region on the right of the $y$ axis. The rectangle looks like the red region in the figure below.
The rectangle is so thin in this diagram that it looks like a line segment. This particular example is for the radius $y = 7.$ The length of the rectangle is therefore $\sqrt{64-7^2} = \sqrt{15} \approx 3.87,$ which is consistent with the diagram. The cylindrical shell we get from this rectangle looks like this:
Now if we start with a strip just above $y = 6$ and then keep adding strips above it until we reach $y = 8,$ we will generate cylindrical shells that fill the volume we were trying to measure. That's why the shell method gives the correct measurement of this volume when the integral is set up as
$$ \int_6^8 2 \pi y\left(\sqrt{64-y^2}\right)\,\mathrm dy. $$
Now suppose we use this integral instead: $$ \int_0^{\sqrt{28}} 2\pi x \left(\sqrt{64-x^2}\right)\,\mathrm dx. $$
The obvious interpretation is that we're integrating across the region from left to right instead of from bottom to top. (Your intuition is correct about that.) So instead of thin rectangles with their long dimensions horizontal, we must be looking at thin rectangles with their long dimensions vertical; the narrow dimension of the rectangle is $\mathrm dx,$ which is measured along the $x$ axis.
Now we need the rectangle to have a length and the cylinder to have a radius. The only way a thin rectangle with its long dimension vertical is going to make a cylindrical shell is if we rotate it around a vertical axis, and the obvious choice is the $x$-axis because then $x$ is the distance of the rectangle from that axis. That means $\sqrt{64-x^2}$ is the length of the rectangle.
Note that we are integrating $x$ from $0$ to $\sqrt{28}.$ For example, therefore, one of the thin rectangles will be at $x = 1.$ The length of this strip is $\sqrt{64-1^2} = \sqrt{63} \approx 7.94.$ Where in the region shown in the first figure are we going to fit a rectangle that is about $7.94$ units long?
The answer is, there is no such rectangle inside that region. The formula doesn't apply to that region at all. Instead, a reasonable interpretation is that the rectangle at $x = 1$ is the red region in the figure below, where the gray region is the region we will be covering with the rest of the thin rectangles as $x$ goes from $0$ to $\sqrt{28}.$
The cylinder of radius $x$ that we get by rotating this rectangle is
When we put together all of the cylindrical shells between $x = 0$ and $x = \sqrt{28},$ they fill a solid object that looks like this:
So this is a solid of revolution whose volume is measured by the integral $$ \int_0^{\sqrt{28}} 2\pi x \left(\sqrt{64-x^2}\right)\,\mathrm dx. $$
You may have had some other idea in mind for how that integral would represent the volume of an object, but aside from the question of whether that idea made any sense, there should be no doubt that if we wanted to measure the volume of the object above by the shell method, the integral $$ \int_0^{\sqrt{28}} 2\pi x \left(\sqrt{64-x^2}\right)\,\mathrm dx $$ is the integral that we would naturally want to use. So unless the volume of that object is coincidentally the same as the volume of the object that you're actually supposed to be measuring, you won't get the right answer. And you really should not rely on lucky coincidences in your calculations.
Think of the shell method as a machine that helps you get answers, but it doesn't work unless you feed the right input into the right places in the machine. For example, a car will run well and take you on a trip if you pour coolant into the radiator and gasoline into the fuel tank, but if you put the gasoline in the radiator and the coolant in the fuel tank you will not get where you wanted to go.
Eventually, if you get very good at this, you should be able to build your own machines to compute volumes of solids of rotations, but until then, read the user manual carefully (it should be in your textbook) and use the machine you were given exactly as the manual says.
The following notes are for the benefit of pedantic readers who might object to some of the details above on technical mathematical grounds. I don't think you need to understand these things in order to see why you should use the shell formula (in this case) by integrating over $y$ rather than over $x.$
[note 1] In the context of standard calculus, I'm being a little non-rigorous here by using $\mathrm dy$ as the width of a rectangle; if you want, you can replace $\mathrm dy$ with a small real number $\Delta y,$ similar to the way you make a rectangle in a Riemann sum.
[note 2] Depending on how you define the radius of the cylindrical shell, this volume formula may be approximate, although it is accurate in the limit as $\mathrm dy$ goes to zero. If you make $y$ be exactly halfway between the inner and outer radius of the shell, the formula is exact.
[note 3] Again I'm a little non-rigorous in how I add up the volumes of a set of shells. If you were in a rigorous calculus course you might do a kind of Riemann sum of the shells and let $\Delta y$ go to zero to prove the formula.