Is there a solution to $y=\ln(x)+x$ which yields an answer in the form $x^2=...$

Question

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:

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.

2. 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.

Answer 1

$$y=\ln(x)+x \implies e^y=xe^x \implies x=W(e^y) \implies x^2=(W(e^y))^2 $$ 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

Answer 2

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.

Answer 3

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 $\int x^2\,\mathrm dy =\int y\,\mathrm dy$.