Is there any easy way to decide which method I should use to find the volume of a revolution in Calculus? I'm currently in the middle of my second attempt at Calculus II, and I am getting tripped up once again by this concept, which seems like it should be rather straight forward, but I can't figure it out. If I have the option of the disk/washer method and the shell method, how do I know which one I should use?
$\begingroup$
$\endgroup$
2
asked Jun 28, 2012 at 17:02
-
$\begingroup$ After a while, if you do enough examples, you will in essence have "seen everything." $\endgroup$– André NicolasJun 28, 2012 at 17:26
-
$\begingroup$ Best criterion: what integral is easier? $\endgroup$– Martín-Blas Pérez PinillaMar 27, 2018 at 9:41
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
1
The first thing to understand is that you don’t directly choose the method of integration: you determine what kind of integration will be easier, based on the shape of the region in question, and that determines which method you’ll use.
Draw the region that’s being revolved. Then ask yourself: does it slice up nicely into vertical strips, or do horizontal strips work better?
If the region has boundaries of the form $y=f(x)$, $y=g(x)$, $x=a$, and $x=b$, the answer is almost always that vertical strips are simpler: for each $x$ from $a$ to $b$ you have a strip of length $f(x)-g(x)$ or $g(x)-f(x)$, depending on which of $f(x)$ and $g(x)$ is larger.
Similarly, if the region has boundaries of the form $x=f(y)$, $x=g(y)$, $y=a$, and $y=b$, the answer is almost always that horizontal strips are simpler: for each $y$ from $a$ to $b$ you have a strip of length $f(y)-g(y)$ or $g(y)-f(y)$, depending on which of $f(y)$ and $g(y)$ is larger.
The case of a region bounded by $y=f(x)$ and $y=g(x)$, where you have to solve for the points of intersection of the two curves in order to find the horizontal extent of the region, is really just special case of (1). Similarly, if the region is bounded by $x=f(y)$ and $x=g(y)$, and you have to solve for the vertical extent of the region, you’re looking at a special case of (2).
What you want is a way of slicing the region into vertical or horizontal strips whose endpoints are defined in the same way. Take, for instance, the region bounded by $y=x$ and $y=x(x-1)$. If you slice it into vertical strips, each strip runs from the parabola at the bottom to the straight line at the top, so the strip at each $x$ between $0$ and $2$ has its bottom end at $x(x-1)$ and its top end at $x$. If you were to slice it into horizontal strips, the ones between $y=0$ and $y=2$ would have their left ends on the straight line and their right ends on the parabola, but the ones between $y=-1/4$ and $y=0$ would have both ends on the parabola. Thus, you’d need a different calculation to handle the part of the region below the $x$-axis from the one that you’d need for the part above the $x$-axis.
Whether the boundaries are given in the form $y=f(x)$ or the form $x=f(y)$ is often a good indication: the former tends to go with vertical slices and the latter with horizontal slices. It’s far from infallible, however, and in some problems some boundaries are given in one form and some in the other. You should always look at a picture of the region. You want slices whose endpoints are defined consistently, and you want slices that don’t have any gaps in them.
Once you’ve decided which way to slice up the region, sketch in the axis of revolution. If it’s parallel to your slices, each slice will trace out a cylindrical shell as it revolves about the axis. If, on the other hand, it’s perpendicular to your slices, each slice will trace out a washer or disk as it revolves about the axis. In either case the proper method of integration has automatically been determined for you.
answered Jun 28, 2012 at 17:21
-
$\begingroup$ "If it’s parallel to your slices, each slice will trace out a cylindrical shell as it revolves about the axis. If, on the other hand, it’s perpendicular to your slices, each slice will trace out a washer or disk as it revolves about the axis." Just wanted to point out that this is an absolutely key fact that is rarely stated outright, thank you! (i.e. once a choice of rectangular strips is made, the method is entirely determined by whether the strip is parallel or perpendicular to the axis of rotation.) $\endgroup$ Sep 12, 2018 at 5:35
$\begingroup$
$\endgroup$
5
Use whichever method you find easiest to apply for each problem you encounter. They should give the same result (or something is wrong with the way you apply them).
As a practical matter, just pick one of them at random until you work up enough experience to have a hunch about what is likely to work easiest for the kind of shape you're looking at. If your first choice gets you into integrals that are hard for you to solve, put it aside for a while and see if it the other way is easier going.
Continue switching between the two methods, throwing progressively more effort at each until one of them yields.
answered Jun 28, 2012 at 17:12
hmakholm left over Monicahmakholm left over Monica
284k24 gold badges425 silver badges685 bronze badges
-
1$\begingroup$ Also, setup a few problems using both methods to see how one way can be easier than the other. $\endgroup$– dlsJun 28, 2012 at 17:13
-
$\begingroup$ This is the method I used to use last time I took Calc II. If I didn't know which one is was, I would do both methods until each got pretty hard, then I would pick the easier looking equation to finish. But sometimes it would then end up as the one that looked easier in the middle was harder to finish. I guess it is kind of just random then. $\endgroup$ Jun 28, 2012 at 17:19
-
$\begingroup$ @Nick: It is absolutely not random. There is a straightforward way to choose that works in the overwhelming majority of problems; see my answer. $\endgroup$ Jun 28, 2012 at 17:23
-
$\begingroup$ @Brian: I assumed we were dealing with cases where both ways of slicing make sense in the first place. $\endgroup$ Jun 28, 2012 at 17:29
-
$\begingroup$ @Henning: I’ve the advantage of having spent many years watching people struggle with volumes of revolution, so I knew that that wasn’t a safe assumption. Textbooks often contribute to the problem, both by giving the impression that one chooses the method first and by including exercises requiring students to use the ‘wrong’ method. $\endgroup$ Jun 28, 2012 at 17:35
$\begingroup$
$\endgroup$
I think that there are some standard formulae in integral calculus to calculate the Volume of revolution. These offer you a straight forward method ( I don't know anything about the dish washer method ) but you still have to know how to trace the curves to apply the following formulae on them:-
$V=\pi\int_a^by^2dx$ where $y=f(x)$ is the equation of the curve and $x=a$ and $x=b$ are the limits of the revolution.
$V=\pi\int_c^dx^2dy$ where $x=g(y)$ is the equation of the curve and $y=c$ and $y=d$ are the limits of the revolution.
To calculate the volume of revolution when polar coordinated are given, you substitute $y=cos\theta$ and $x=sin\theta$ and calculate the limits accordingly. Also, you need to see that the given formulae are valid only when the revolution is occuring about the x-axis or y-axis.If the curve is revolving around x=constant or y=constant, then you need to change the origin and make suitable adjustements.
