How quickly does the water increases in the tank reaches the water level is 1m? I have a body with the following parameters:
$y = 0, y = \sqrt{x − 1}, y = 2$ and $x = 0$
On the y-axis. And it is rotated
The question states: The body is a water tank with the height of 2m that is being filled with water at a constant speed of $\frac{1}{2}m^3/min$. How quickly does the water increases in the tank reaches the water level is 1m?
 A: Solving for x, the equation $y = \sqrt{x - 1}$ at y = 1, gives x = 2. At $1$m depth the surface area of the water is $\pi(2)^2 = 4\pi\ m^2$. Hence the rate of  the level rising in the tank is $\frac{0.5}{4\pi} = .0397887$ m/min. 
I may get down votes for this solution as it bypasses calculus. But I suggest you pursue the calculus method to help with your studies. Keep this solution in mind when looking for a faster method. That is, tank filling and balloon inflation rates can be determined by dividing the rate of filling by the surface area at a specified fluid level or inflation size.
This is the calculus method:
Because our volume is obtained by revolving $y = \sqrt{x-1}$ around the $y$ axis, the equation needs to be transformed to $x = y^2 + 1$.
$V = \int_0^y \pi\cdot (y^2 + 1)^2 dy$
$\frac{dV}{dy} = \pi\cdot (y^2 + 1)^2$
$\frac{dV}{dt} = \frac{dV}{dy} \cdot \frac{dy}{dt}$
When $y = 1$
$0.5 = \pi\cdot (1^2 + 1)^2\cdot \frac{dy}{dt}$
$\frac{dy}{dt} = \frac{0.5}{4\pi} = .0397887$ m/min
A: There are numerous problems of this kind on this site copied from various
exercises in homework or textbooks, describing a tank of some particular shape,
saying that water enters the tank at a certain rate of volume per unit time,
and asking how fast the water level in the tank is increasing when the
water is at a certain level in the tank.
The "obvious" way to work these problems is to compute the volume of the water as a function of the level it has reached, then differentiate this volume relative to the variable that tells the level of the water. Using this derivative, $dV/dh,$
and the given inflow of water, $dV/dt,$ you can easily compute $dh/dt,$
which is what you were asked to compute.
The trick is to understand that in any problem of this kind,
$$V(h) = \int_{h_0}^h A(\eta) d\eta,$$ where $V(h)$ is the volume of water when the level is $h,$ where $A(\eta)$ is the surface area of the top of the water when the level is $\eta,$ and where $h_0$ is the level at the bottom of the tank.
(We usually set things up so that $h_0 = 0,$ but as I'm about to show, this is generally irrelevant.)
Applying the fundamental theorem of calculus,
we can conclude that $$\frac{d}{dh} V(h) = A(h).$$
It's often harder to compute $V(h)$ than $A(h)$--in this case you have to compute $A(h)$ and then integrate it in order to get $V(h)$--and on top of that the "obvious" method then forces you to differentiate $V(h)$.
So you integrate $A(h)$ and then differentiate the result, getting back $A(h)$;
what a waste of effort!
In the easy approach, we just use the fact that $dV/dh = A(h),$ compute $A(h),$ and get a simple answer like Phil H's.
