Integration limits when integrating both sides

I have been working on solving differential equations and this is really cracking me up. I obtained the following equation:

dz/dr = r

and I wish to obtain z in terms of r, given that we know that z = 0 when r = 0; is this:

$dz = r dr$

$\int_0^z 1\,dz = \int_0^r r \,dr$

$==> z = r^2/2$

the right way of doing this(are the limits of the integration correct)?

thank you

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That's right. And it's the simplest way to do these initial value type questions. – coffeemath Nov 11 '12 at 2:29

It may be more illustrative to do the problem in as slightly different way to see what's going on. In this context, $z$ is shorthand for $f(r)$, where $f$ is some function. $\frac{\mathrm{d}z}{\mathrm{d}r}=r$ really means $f'(r)=r$. Then we can apply $\int_0^R\cdot\mathrm{d}r$ to both sides of that equation: $\int_0^Rf'(r)\mathrm{d}r=\int_0^R r\mathrm{d}r$. On the left hand side we have $f(R)-f(0)$ by the fundamental theorem of calculus, and on the right hand side we have $R^2/2$. Finally, since we know $f(0)=0$ (since $z=0$ when $r=0$), this just says $f(R)=R^2/2$. This should work for any $R$, so we can write $f(r)=r^2/2$ or just $z=r^2/2$.
This is essentially the same calculation you did, but be careful, your original calculation contained two errors in notation that cancelled out. If you write $\int_0^z 1\mathrm{d}z$, it should evaluate to something like "the value of $z$ at $r=$'$z$' minus the value of $z$ at $r=0$ (which happens to be $0$)", which isn't quite what you want. By writing $z$ on the next line instead of something more like $z\mid_{r=z}$ you made a second mistake (in notation) that fixed the issue.