# Trying to figure out where the numbers are coming from for this work/integration problem

There is a round container pushing its side loaded entirely with fluid. The size of the container is 4ft, and also the size of the container is 10 ft.

Here is a dreadful illustration of it yet close adequate:

I need to set up an indispensable to locate the job called for to pump all the fluid to a degree 1 feet over the top of the container.

So I have my formula:

Work = π (weight thickness) ∫ (location) (variation) (density)

Someone else did this trouble and also got a bounds of 2 to - 2, a variation of (3 - y), a location of 2x, and also a density of dy. Where did the 2 to - 2, (3 - y), and also 2x originate from? This is possibly instinctive yet I do not see it.

(Where did $\pi$ originated from?)

Consider the container right on, to make sure that you just see the round side. Visualize that circle on the aircraft, with the facility at $(0,0)$. Given that the circle has size $4$, all-time low of the circle goes to $(0,-2)$, and also the top of the circle goes to $(0,2)$. You intend to raise the fluid to one foot over the top of the container, so you are attempting to raise it to the line $y=3$.

Currently, take a straight piece of that circle of density $\Delta y$, at elevation $y_i$, where $y_i$ is someplace in between $-2$ and also $2$. This piece represents a quantity of fluid that is about equivalent to the location of this piece, times $10$ (to make up the size of the container).

To locate the location, note that the circle has formula $x^2+y^2 = 4$, to make sure that the $x$ coordinate is offered by $x=\pm\sqrt{4-y^2}$. So the size of the piece is mosting likely to be $2x=2\sqrt{4-(y_i)^2}$. The density is $\Delta y$. So the quantity of the piece of fluid is about equivalent to $$20\sqrt{4 - y_i^2}\Delta y\text{ cubic feet.}$$ Now, you intend to raise all of it the means to $y=3$ ; just how much from $y=3$ are you? Given that the piece is at elevation $y_i$, the range to $y=3$ is $3-y_i$.

So the job carried out in raising this set piece of fluid is about: $$\begin{align*} \text{Work}&=\text{Weight}\times\text{Displacement}\\ &= \left(\text{Volume}\times\text{Density}\right)\times\text{Displacement}\\ &\approx \left(20\sqrt{4-y_i^2}\Delta y\right)\times\text{Density}\times(3-y_i)\\ &= \text{Density}\times 20(3-y_i)\sqrt{4-y_i^2}\Delta y. \end{align*}$$ (Be certain to get the devices right ; the thickness need to remain in extra pounds per cubic feet, in which instance the devices of job will certainly be ft - pounds.)

Currently, to get the complete job, you slice the container right into these slim pieces, identify the benefit each, and also add all of it up: $$\text{Work}\approx \sum_{i=1}^n \text{Density}\times 20(3-y_i)\sqrt{4-y_i^2}\Delta y.$$ Taking the restriction as $n\to\infty$, this comes to be an indispensable. The integrand is $$\text{Density}\times 20(3-y)\sqrt{4-y^2}\,dy.$$ (The $\sqrt{4-y^2}$ is the value of $x$ at the piece).

What are the restrictions of assimilation? Returning to the image, what are the feasible values of $y$? Due to the fact that $y$ represents where you remain in the container, and also the container expands from $y=-2$ to $y=2$, after that the restrictions of assimilation go from $-2$ to $2$.

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