Rotating Parabolas

The volume of the solid up to an arbitrary height h when rotated around the y axis is given by the volume of revolution formula. $V=\int_{0}^{h}\pi x^{2} dy=\int_{0}^{h}\pi y dy$ Evaluating this we find $V=\frac{1}{2}\pi h^2$ We also know that $\frac{dV}{dt}=2\pi$ If we integrate this we can find a function for the volume at a time t. $V=2\pi t$ We can now equate these two equations for volume to find a function for h at a time t. $V=2\pi t=\frac{1}{2}\pi h^2$ We rearrange to find $h=2\sqrt {t}$ We can differentiate this to find how the height h is varying with time t. $\frac{dh}{dt}=\frac{1}{\sqrt {t}}$ We need to now find the time at which the volume is equal to 8 pi. Using our equation for V that is a function of time. $8\pi=2\pi t$ $t=4$ We can now use our rate of change of h to find the answer $\frac{a}{b}=\frac{dh}{dt}_{t=4}=\frac{1}{\sqrt{4}}=\frac{1}{2}$ So $a+b=3$

$y$ is the height of water in the structure, while $x$ is the radius of a circular slice of the structure at height $y$ . Since we are only dealing with positive $y$ values, we can define $x=\sqrt{y}$ .

The area of a slice of the structure at height $y$ is $A=\pi x^2=\pi y$ .

Since the volume is a sum of areas times a small change in height:

$V=\int A \ dy=\int \pi y \ dy=\frac{1}{2}\pi y^2$

Plugging in $V=8 \pi$ :

$8 \pi=\frac{1}{2}\pi y^2 \implies y=4$

By the chain rule:

$\frac{dV}{dt}=\frac{dV}{dy}*\frac{dy}{dt}$

We know that $\frac{dV}{dt}=2\pi$ and that $\frac{dV}{dy}=A=\pi y=4\pi$ , so:

$2\pi=4\pi \frac{dy}{dt} \implies \frac{dy}{dt}=\frac{1}{2}=\frac{a}{b}$

Therefore, the answer is $a+b=1+2=\fbox{3}$ .

when the water depth is $a$ , the volume is :

$v = \int_{0}^{a} \pi x^2 dy$

$= \int_{0}^{a} \pi y dy$

$= \frac{\pi}{2} a^2$

$\frac{dv}{dt} = \pi a \frac{da}{dt}$

when $v=8\pi, a=4$ , so since $\frac{dv}{dt}=2\pi,$

$2 \pi = \pi (4) \frac{da}{dt}$

$\frac{da}{dt} = \frac{1}{2}$

$1+2=\boxed3$

For $y = {x^2}$ parabola rotated around y-axis, the equation to find volume of the solid formed is: $\int_{a}^b pi {x^2} dy$

Calculating the height of solid when 8pi ${units^3}$ of water is filled. i.e. 8pi = $\int_{0}^y pi {x^2} dy$ where dy = 2xdx hence,

8pi = $\int_{0}^y 2pi {x^3} dx$

calculating $\boxed{y = 2}$ from above equation.

Since, 2pi ${units^3}$ per sec in flow rate,

hence new volume would be 10pi ${units^3}$

On carrying out same procedure to calculate w

10pi = $\int_{2}^w 2pi {x^3} dx$

$\boxed{w = 2.5}$

Now change in height due to increase in water level would be

$w - y$ ,

i.e. 2.5 - 2 = 0.5 = $\frac{1}{2}$

hence a=1 and b=2 ,

a+b = $\boxed{3}$

We begin by finding the volume of the figure, given its height $h$ measured from the x-axis. The line $y=h$ intersects the parabola $y=x^{2}$ when $x=\sqrt{h}$ . Applying the shell method for finding volume, the volume as a function of the height can be found:

V=2\pi \int\limits_0^\sqrt{h} x^{2}= 2\pi \left[\frac{x^{4}}{4}\right]_0^\sqrt{h} =\frac{\pi}{2}h^{2} .

Given that the current volume is $8\pi$ , plugging this value into the above equation yields $h=4$ . We then take the derivative with respect to time of the above equation to solve for $\frac{\mathrm{d}h}{\mathrm{d}t}$ (Plugging in $2\pi$ for $\frac{\mathrm{d}V}{\mathrm{d}t}$ and $4$ for $h$ ):

$\frac{\mathrm{d}V}{\mathrm{d}t}=\pi h\frac{\mathrm{d}h}{\mathrm{d}t} \Rightarrow 2\pi=\pi (4) \frac{\mathrm{d}h}{\mathrm{d}t} \Rightarrow \frac{\mathrm{d}h}{\mathrm{d}t}=\frac{1}{2}$ .

Therefore, $\frac{1}{2}=\frac{a}{b}$ and our answer is $a+b=1+2=\boxed{3}$ .