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  1. How could I write $y=\ln(x)+x$ as $x^2=...$

Since there might be another solution to this problem I'll give some background. So I had a math test yesterday where they wanted you to calculate the volume of (V) when turned around the y-axis, see here:

enter image description here

The formula for this is pretty easy: $\pi r^2h-\int_a^b(x)^2 dy$

The notation might be different (Dutch) so h is the height of a cylinder, and $\int_a^b(x)^2 dy$ means the integral of the primitive of $(x)^2$.

Since the formula needs $x^2$ it has to be written in that way.

  1. The math test was a pretty big bummer, even more so since I learned hard for it and understood all of the higher concepts but I (and many others) stranded on the basic things like this.

Normally they make the concepts more complicated so you have to combine multiple, however this time there were just a lot of hard things like the above, writing $y=\ln(x)+x$ as $x^2=...$. They don't seem to require a lot of insight more so knowing the rules. Particularly if the book and your teacher don't even explain what things like $\ln$ and e actually mean.

Is there a way to learn solving these problems which require rules with understanding when your high school teacher/ book doesn't tell you about it? I really like trying to understand math but this seems more like just learning the rules. Is there a way to mix these two together? For instance a book on mixing high school calculus together with a deeper insight.

This is my first stackexchange post so I hope it's fine, I couldn't find anything about asking multiple questions at the same time so I hope it's allowed.

I also couldn't find anything about these the required level of math so I hope my high school math is allowed, if not I'd still like to know the answer to my second question. Which I think is more important and would allow me to enjoy math more anyway. That's why I like to look at stackexchange, things don't just get answered; insight is provided.

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  • $\begingroup$ Do you want to write $x^2$ in terms of $y$? $\endgroup$
    – Jakobian
    May 30, 2018 at 20:00
  • $\begingroup$ There are other ways to compute $\int_a^b x^2dy$. Perhaps the intent of the question was to solve it that way? $\endgroup$
    – Michael Burr
    May 30, 2018 at 20:01
  • $\begingroup$ I am not sure how to write $lnx+x$ in terms of $x^2=$, but can you tell me which cito exam this happens to be? (link) $\endgroup$
    – imranfat
    May 30, 2018 at 20:02
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    $\begingroup$ Where does your $\ln x $ come from? The linked image looks like the standard parabola to me $\endgroup$
    – Hagen von Eitzen
    May 30, 2018 at 20:06
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    $\begingroup$ @imranfat Ik denk dat je helemaal gelijk hebt, dat is wat ik in ieder geval heb geleerd. De opgave zal wel meer context hebben gehad die ik gemist heb. Er is een herkansing dus dan kan ik daar op letten. Overigens was het gemiddelde een 4,8/10 dan lijkt er sowieso al iets niet te kloppen. Dat soort dingen gebeuren op de middelbare vrijwel nooit tenzij toetsen te moeilijk zijn. $\endgroup$
    – SanderSantema
    May 31, 2018 at 5:07

3 Answers 3

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As I read your question, there is the relationship $y=\ln(x)+x$ and you want to calculate $$ \int_a^b x^2\,dy. $$ One way to solve this would be to find a formula for $x^2$ in terms of $y$ and integrate. This seems hard because it isn't obvious how to solve for $x^2$ in terms of $y$ (and if it is even possible in elementary functions).

Alternately, you can turn the $dy$ into a $dx$. Since $y=\ln(x)+x$, it follows that $$ dy=\frac{1}{x}dx+dx. $$ Substituting this into the formula gives $$ \int_{y=a}^{y=b}x^2\left(\frac{1}{x}dx+dx\right)=\int_{y=a}^{y=b}x\,dx+\int_{y=a}^{y=b}x^2\,dx. $$ This, we can integrate to $$ \left.\left(\frac{1}{2}x+\frac{1}{3}x^3\right)\right|_{y=a}^{y=b}. $$ Now, if $a$ and $b$ are "nice," such as $y=1$ or $y=1+e^2$, then you can solve for $x$ and substitute.

As @HagenVonEitzen mentions, the given formula doesn't seem to match the problem described by the OP. There may have been an error in the original set-up, leading to a much harder problem to solve.

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  • $\begingroup$ I'm trying to understand why $\frac{1}{x}dx+dx$ can be filled in after $x^2$. Could you clarify why it's the same? I might be possibly missing something since in our book and practice problems the writing of dx and dy is only a formality after the formula. They don't explain what it means. The graph was only to illustrate the problem. $\endgroup$
    – SanderSantema
    May 30, 2018 at 20:23
  • $\begingroup$ In the line above, we found that $dy=\left(\frac{1}{x}+1\right)dx$, so it's a direct substitution at that point. $\endgroup$
    – Michael Burr
    May 30, 2018 at 20:30
  • $\begingroup$ Ah that's a lack of explanation in the book, or different notation since dy only means you're doing something with the y-axis in my book. Therefore it isn't something which I knew could be substituted, that might be it. Would you just fill in the values of y=a and y=b in X like you'd do normally? $\endgroup$
    – SanderSantema
    May 30, 2018 at 21:00
  • $\begingroup$ By $y=a$, we mean that we substitute the solution to the equation $\ln(x)+x=a$ for $x$. The same goes for $b$. This is hard to solve. You could say that the inverse function of $f(x)=\ln(x)+x$ is $\theta(x)$, for example. We know the function has an inverse, because it's one-to-one (it's strictly increasing). Then we can write our solution in the terms of $\theta$. This is almost the same approach as i used in my solution. $\endgroup$
    – Jakobian
    May 30, 2018 at 21:07
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$$y=\ln(x)+x \implies e^y=xe^x \implies x=W(e^y) \implies x^2=(W(e^y))^2 $$ Where $W$ is W Lambert function. As you see, this is involves a special function, so it's probably not the intented way to go

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  • $\begingroup$ If this is right, unfortunately I can't verify it since I don't know Lambert's function the problem is solved. At least as far as I've given context, since I didn't know if this could be solved without using something like Lambert's function or high school pre-calculus. Apparently context is missing. $\endgroup$
    – SanderSantema
    May 30, 2018 at 20:46
  • $\begingroup$ This seems to be the answer to the question in the title although probably not the answer in my specific context. The question in my specific context is missing details, I didn't know they were missing. Therefore the question can't be solved in my context. It might still be valuable for people who take the wrong approach and try to do the same as I did. $\endgroup$
    – SanderSantema
    May 30, 2018 at 20:57
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From $y=x^2$, we see $x=\sqrt y$. Hence the volume is $$\int_0^4(\pi2^2-\pi\sqrt y^2)\,\mathrm dy=\pi \int_0^4 (4-y)\,\mathrm dy,$$ which you should find quite tractable.

Or in fact as $y=x^2$, you get immediately that $\int x^2\,\mathrm dy =\int y\,\mathrm dy$.

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  • $\begingroup$ Perhaps you should add a caveat about your interpretation of the problem. As written, it's not quite related to the OP's question. $\endgroup$
    – Michael Burr
    May 30, 2018 at 20:10
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    $\begingroup$ @MichaelBurr Actually, I am afraid that the OP's problem seems to be a case of xy problem (and even somewhat literally) $\endgroup$
    – Hagen von Eitzen
    May 30, 2018 at 20:12
  • $\begingroup$ No @MichaelBurr is right. It is specifically about writing $y=\ln(x)x$ as $x^2$. The practice problems were problems like xy but somewhat harder, the discrepancy between these problems and the actual test was the problem. $\endgroup$
    – SanderSantema
    May 30, 2018 at 20:30
  • $\begingroup$ XY in this context means that you're asking about a part of the problem, but you've already gone off-track earlier in the problem. $\endgroup$
    – Michael Burr
    May 30, 2018 at 20:31
  • $\begingroup$ @MichaelBurr XY thanks I understand now. $\endgroup$
    – SanderSantema
    May 30, 2018 at 20:52

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