John's Calculus Integration w/ Simpson's Rule

[MarkFL]

Here is the question:

Calculus Integration Problem?

y = tan−1 x between x = 0 and x = 1 is rotated around the y-axis to form a container. The container is filled with water. Use n = 6 subintervals and Simpson's rule to approximate the work required to pump all of the water out over the side of the container. Give your answer in decimal form.
(Distance is measured in meters, the density of water is 1000 kg/m3, and use 9.8 m/s2 for g.)

I know that I should use cylindrical shells as it is rotated by the y-axis. But beyond that I' am having trouble setting up the integral and evaluating it with Simpson's.

Thanks for the help in advance!

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello John,

If you were trying to compute the volume of the resulting tank, then either the shell or disk method would be appropriate, but we are computing the work done to empty the tank, so we want to decompose the volume into disks, since each disk will have all parts at the same gravitational potential energy.

Let's first take a look at a cross-section of the tank through the axis of symmetry:

View attachment 1153

The radius $r$ of an arbitrary slice I have drawn in red and the distance $d$ through which it must be lifted is drawn in green. We know the amount of work required to lift this slice is the product of the applied force and the distance over which this force is applied, hence:

\(\displaystyle dW=Fd\)

Now, the applied force is simply the weight of the slice:

\(\displaystyle F=mg\)

and the mass of the slice is the product of its mass density $\rho$ and it volume $V$:

\(\displaystyle m=\rho V\)

and of course the volume of the slice is:

\(\displaystyle V=\pi r^2\,dy\)

where the radius of the slice is:

\(\displaystyle r=x=\tan(y)\)

Now, putting this together, we find:

\(\displaystyle dW=Fd=(mg)\left(\frac{\pi}{4}-y \right)=(g\rho V)\left(\frac{\pi}{4}-y \right)=g\rho\left(\pi r^2\,dy \right)\left(\frac{\pi}{4}-y \right)=\pi g\rho\tan^2(y)\left(\frac{\pi}{4}-y \right)\,dy\)

Summing up all the elements of work, we obtain:

\(\displaystyle W=\pi g\rho\int_0^{\frac{\pi}{4}}\tan^2(y)\left( \frac{\pi}{4}-y \right)\,dy\)

Applying Simpson's Rule with $n=6$ on the integral, where:

\(\displaystyle f(y)=\tan^2(y)\left( \frac{\pi}{4}-y \right)\)

and:

\(\displaystyle y_k=\frac{\frac{\pi}{4}-0}{6}k=\frac{\pi}{24}k\) with \(\displaystyle k\in\{0,1,2,3,4,5,6\}\)

we find:

\(\displaystyle S_6=\frac{\frac{\pi}{4}-0}{3(6)}\left[f\left(y_0 \right)+4f\left(y_1 \right)+2f\left(y_2 \right)+4f\left(y_3 \right)+2f\left(y_4 \right)+4f\left(y_5 \right)+f\left(y_6 \right) \right]\)

\(\displaystyle S_6=\frac{\pi}{72}\left[f\left(y_0 \right)+4f\left(y_1 \right)+2f\left(y_2 \right)+4f\left(y_3 \right)+2f\left(y_4 \right)+4f\left(y_5 \right)+f\left(y_6 \right) \right]\)

\(\displaystyle f\left(y_0 \right)=f\left(\frac{\pi}{24}\cdot0 \right)=f(0)=\tan^2(0)\left( \frac{\pi}{4}-0 \right)=0\)

\(\displaystyle 4f\left(y_1 \right)=4f\left(\frac{\pi}{24}\cdot1 \right)=4f\left(\frac{\pi}{24} \right)=4\tan^2\left(\frac{\pi}{24} \right)\left( \frac{\pi}{4}-\frac{\pi}{24} \right)\approx0.045376065047798\)

\(\displaystyle 2f\left(y_2 \right)=2f\left(\frac{\pi}{24}\cdot2 \right)=2f\left(\frac{\pi}{12} \right)=2\tan^2\left(\frac{\pi}{12} \right)\left( \frac{\pi}{4}-\frac{\pi}{12} \right)\approx0.075185401439313\)

\(\displaystyle 4f\left(y_3 \right)=4f\left(\frac{\pi}{24}\cdot3 \right)=4f\left(\frac{\pi}{8} \right)=4\tan^2\left(\frac{\pi}{8} \right)\left( \frac{\pi}{4}-\frac{\pi}{8} \right)\approx0.269506042226324\)

\(\displaystyle 2f\left(y_4 \right)=2f\left(\frac{\pi}{24}\cdot4 \right)=2f\left(\frac{\pi}{6} \right)=2\tan^2\left(\frac{\pi}{6} \right)\left( \frac{\pi}{4}-\frac{\pi}{6} \right)\approx0.174532925199433\)

\(\displaystyle 4f\left(y_5 \right)=2f\left(\frac{\pi}{24}\cdot5 \right)=4f\left(\frac{5\pi}{24} \right)=4\tan^2\left(\frac{5\pi}{24} \right)\left( \frac{\pi}{4}-\frac{5\pi}{24} \right)\approx0.308290092997036\)

\(\displaystyle f\left(y_6 \right)=f\left(\frac{\pi}{24}\cdot6 \right)=f\left(\frac{\pi}{4} \right)=\tan^2\left(\frac{\pi}{4} \right)\left( \frac{\pi}{4}-\frac{\pi}{4} \right)=0\)

Adding these together, we find:

\(\displaystyle S_6\approx0.0380870342601150\)

For comparison, a numeric integration function returns:

\(\displaystyle 0.038148452745930\)

Now, using the approximation we obtained, we may state:

\(\displaystyle W\approx\pi g\rho\cdot0.0380870342601150\)

Using the given data:

\(\displaystyle g=9.8\frac{\text{m}}{\text{s}^2},\,\rho=1000\frac{\text{kg}}{\text{m}^3}\)

and observing the integral has units of $\text{m}^4$, we have:

\(\displaystyle W\approx\pi\left(9.8\frac{\text{m}}{\text{s}^2} \right)\left(1000\frac{\text{kg}}{\text{m}^3} \right)\cdot0.0380870342601150\text{ m}^4\approx1172.60868088\text{ J}\)

For comparison, the exact answer is:

\(\displaystyle W=\frac{1225\pi}{4}\left(16\ln(2)-\pi^2 \right)\text{ J}\approx1174.5\text{ J}\)