Work required to empty water out of a vertical ellipsoid tank.

[Pull and Twist]

This is another problem I am having difficulty with... I set it up like I've been working the book problems, especially the sphere problems, but can't seem to get the right answer. I feel that I am calculating the radius incorrectly.

I know I am supposed to us \(\displaystyle {x}^{2}+{y}^{2}={r}^{2}\) and \(\displaystyle r=1\) but for the \(\displaystyle y\) value do I take the disk's position from the top... \(\displaystyle 4-y\) or should I be looking at the disk top down to calculate the radius as \(\displaystyle {x}^{2}+{y}^{2}=1\) or even \(\displaystyle {\left(1-x\right)}^{2}+{\left(1-y\right)}^{2}=1\) knowing that the disk shrinks at points along the integral.

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[MarkFL]

I prefer to work problems like this in general terms, and then plug our given data into the resulting formula.

Let's let:

\(\displaystyle w\) = the width of the ellipsoidal tank.

\(\displaystyle h\) = the height of the tank.

\(\displaystyle \ell\) = the length of the spout.

\(\displaystyle \rho\) = the mass density of the fluid.

\(\displaystyle g\) = the acceleration due to gravity.

Now, let's imagine slicing the contents of the tank horizontally into circular sheets. The radius of each sheet will be a function of its distance from the bottom of the tank. So, let's orient a vertical $y$-axis passing through the axis of symmetry of the tank, with its origin at the center of the tank.

To determine the radius of an arbitrary sheet, consider the following ellipse:

\(\displaystyle \frac{4x^2}{h^2}+\frac{4y^2}{w^2}=1\implies r^2=\frac{w^2}{4h^2}\left(h^2-4y^2\right)\)

And so the volume of an arbitrary sheet can be given by:

\(\displaystyle dV=\pi\left(\frac{w^2}{4h^2}\left(h^2-4y^2\right)\right)\,dy\)

Next, we want to determine the weight $u$ of this sheet. Using the definition of weight density, we may state:

\(\displaystyle \rho g=\frac{u}{dV}\,\therefore\,u=\rho g\,dV\)

Next, we observe that the distance $d$ this sheet must be lifted is:

\(\displaystyle d=\frac{h}{2}-y+\ell\)

Thus, using the fact that work is the product of the applied force and the distance through which this force is applied, we find the work done to lift the arbitrary sheet is:

\(\displaystyle dW=ud=\pi\rho g\left(\frac{h}{2}-y+\ell\right)\left(\frac{w^2}{4h^2}\left(h^2-4y^2\right)\right)\,dy\)

And so summing the work elements, we find:

\(\displaystyle W=\frac{\pi w^2\rho g}{4h^2}\int_{-\frac{h}{2}}^{\frac{h}{2}}\left(\frac{h}{2}-y+\ell\right)\left(h^2-4y^2\right)\,dy\)

If we now observe that we may employ the odd and even function rules, this reduces to:

\(\displaystyle W=\frac{\pi w^2\rho g}{4h^2}\left(h+2\ell\right)\int_{0}^{\frac{h}{2}}\left(h^2-4y^2\right)\,dy\)

Evaluating the integral and then simplifying, we get:

\(\displaystyle W=\frac{\pi hw^2\rho g}{12}\left(h+2\ell\right)\)

Plugging in the given data for this problem:

\(\displaystyle h=4,\,w=2,\,\ell=1\)

there results:

\(\displaystyle W=\frac{16\pi\rho g}{12}\left(6\right)=8\pi\rho g\)