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Almost everywhere I look the formula is:

$$ \pi \int_b^a {\left(f(x)^2 - g(x)^2\right) dx} $$

where f(x) is the big function and g(x) is the smaller function.

Though I've run into problems while calculating the volume using this formula, and through trial and error have found that sometimes the correct formula to use is:

$$ \pi \int_b^a {(f(x) - g(x))^2 dx} $$

These are vastly different, and I'm getting quite confused as to which formula is the correct one. The questions that use the second formula are all from my textbook, while my professor uses the first formula during lectures.

Is one of these correct? Are they both correct and just for different questions? Thanks for any help.

RobPratt
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Jackson
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    The first formula is the correct one. If you would like to show an example and your work that leads you to the conclusion that the second one might be correct, we can try to straighten things out. – Doug M Jun 30 '21 at 22:50
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    Your claim, that by trial and error you have determined that the second formula is correct, sounds very suspicious to me, in particular in the absence of any supporting example when the second formula works better. Are you sure? (At any rate, the second formula is not correct, even if it may occasionally produce a correct answer, by coincidence, and I do not expect that to happen very often, only in some very special conditions that virtually never hold. Do not use it.) – Mirko Jul 01 '21 at 00:25

3 Answers3

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The first equation is correct and the second wrong.

enter image description here

Just study this figure where $f(x) - g(x)$ is the same for the red and blue. Your second equation incorrectly says they have the same area.

David G. Stork
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The first formula is correct, subject to confusion that most books (that I use to teach, and that are not my choice) do not clearly explain what are these $f(x)$ and $g(x)$. (Either that, or students try to read too fast and overlook what these $f(x)$ and $g(x)$ are supposed to denote, but at any rate I find the notation $f(x)$ and $g(x)$ unnatural and unfortunate in this context, and no wonder why students would be confused.)

In your question you call these $f(x)$ and $g(x)$ "big function" and "small function" and I suspect you may not quite grasp what these are (though of course I may be wrong). At any rate, I would rather call them "big radius" and "small radius" instead, and denote them say by $R(x)$ (the big radius) and $r(x)$ (the small radius). They depend not only on the functions that describe a bounded region, but also on the axis of revolution. In this notation, the correct formula becomes $$\pi \int_a^b\big(R(x)\big)^2 - \big(r(x)\big)^2 dx$$ (assuming we use washers and the axis of revolution is horizontal). (BTW, why do you have the bounds $a,b$ reversed in your question?)

Say the region is bounded by the curves $y=x^2$ and $y=3x.$ They intersect when $x^2=3x$, that is $x^2-3x=x(x-3)=0$ with solutions $x=0$ and $x=3$, and these are the $a$ and $b$ that are the bounds of integration. Note that when $0<x<3$ we have $x^2<3x$, so $x^2$ is the smaller function (in this interval $[0,3]$) and $y=3x$ is the bigger function, of the two. (Please sketch a picture yourself.)

Now say the axis of revolution is the $x$-axis, note that it has equation $y=0$. To find the big radius $R(x)$ we subtract the $y=0$ (describing the axis of revolution) from the big $y=3x$ to obtain $R(x)=3x-0=3x$. Similarly, to obtain the small radius $r(x)$ we subtract the $y=0$ (describing the axis of revolution) from the small $y=x^2$ to obtain $r(x)=x^2-0=x^2$. The integral is $$\pi \int_0^3 (3x)^2 - (x^2)^2\ dx$$

Next say we consider exactly the same region, but with different axis of revolution, the line $y=-1$. This would change the solid and its volume, and the integral. Again, to find the big radius $R(x)$ we subtract the $y$ describing the axis of revolution from the big $y$, but this time the $y$ describing the axis of revolution is $y=-1$ (not $y=0$), while the big $y$ is just as before, $y=3x$. So we subtract the $y=-1$ from the $y=3x$ to obtain the big radius $R(x)=3x-(-1)=3x+1$. For the small radius we have $r(x)=x^2-(-1)=x^2+1$. The integral is $$\pi \int_0^3 (3x+1)^2 - (x^2+1)^2\ dx$$

Finally say (for the same region) the axis of revolution is the horizontal line $y=9$. Note that the graph of this line is higher up than the graphs of $y=3x$ and $y=x^2$ in the interval $[0,3]$. So, this time to obtain the big radius, we subtract the smaller $y$, namely $y=x^2$ from the $y$ describing the axis of revolution, namely $y=9$. That is, $R(x)=9-x^2$. For the smaller radius we subtract the bigger $y$, namely $y=3x$ from the (even bigger) $y$ decribing the axis of revolution $y=9$, to obtain $r(x)=9-3x$. In the end, the integral is $$\pi \int_0^3 (9-x^2)^2 - (9-3x)^2\ dx$$

Mirko
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just think of the formula $\pi r^2$, you want the area enclosed by: $$r_{inner}\le r\le r_{outer}$$ so it is going to be: $$\underbrace{\pi\int f^2\,dx}_{\text{outer}}-\underbrace{\pi\int g^2\,dx}_{\text{inner}}=\pi\int(f^2-g^2)\,dx$$

Henry Lee
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