# Find the depth of the propane in the tank when it is filled to one-quarter of the tank's volume.

I'm new to this site. My boyfriend has some calculus problems that he's unable to complete due to a family emergency, and so I am trying to help him. His professor hasn't emailed him back about extending the deadline, which is in a few hours. I was hoping someone could help me with this problem since I haven't done calculus in a few years.

A propane tank is in the shape generated by revolving the region enclosed by the right half of the graph of x^2+16y^2=144 and the y-axis about the y-axis. If x and y are measured in meters find the depth of the propane in the tank when it is filled to one-quarter of the tank's volume. (Round to 3 decimal places).

HINT: You will need to use a Graphing Calculator, Online Graphing Calculator, or Computer Algebra System to solve an equation towards the end of your solution to this problem.

I don't know how to begin this problem or how I'd find the integrated value.

• I hate to be that guy, I would recommend reading the chapter. Textbooks at that level are pretty readable and you will certainly have examples that are close enough to this one that should lead you to the answer. Best of luck! – Prototank Jun 30 '16 at 21:05
• I don't have his textbook with me. I've been trying to find problems like this online, but I haven't done calculus in so long that it's hard for me to follow other work. – Rupali Jun 30 '16 at 21:11
• This problem is too ambiguous to solve. Are you filling the shape through the top in the long direction or in the short direction? – Noble Mushtak Jun 30 '16 at 21:15
• I'm not sure. I'm assuming in the long direction. I have 10 tries to get the answer correct though. If you could show me the work for both ways, that would be wonderful. I know that's asking a lot. – Rupali Jun 30 '16 at 21:19
• It is oblate, flat, 12 units semi major along x and 3 unit semi minor axis along y – Narasimham Jun 30 '16 at 21:30

I will solve it vertically. I use sigma notation instead of integral because I am not familiar with manipulating integral form.
Solve $x^2+16y^2=144$ in terms of $x$. $$x=\sqrt{144-16y^2}$$ Since it is rotated around the $y$-axis, we can use the $x$ value as the radius for circle cross sections. The area of these cross sections: $$Area=\pi r^2$$ $$Area=\pi(\sqrt{144-16y^2})^2$$ $$Area=\pi(144-16y^2)$$In the sigma I will use $\frac{i}{n}$ which will give fractions representing all numbers between $0$ and $1$ as it loops through. $y$ must represent values from $-3$ to $3$. We can make the midpoint of the intervals equal and then multiply by a factor. The midpoint of $[-3, 3]$ is $0$. The midpoint of $[0, 1]$ is $\frac{1}{2}$. Subtract $\frac{1}{2}$ from the interval to make $[-\frac{1}{2},\frac{1}{2}]$. The factor is $6$. $\frac{3}{1/2}=6$ $$y=6(\frac{i}{n}-\frac{1}{2})$$ $$y=\frac{6i}{n}-3$$ $\frac{6}{n}$ acts as $dy$ would in integral form. I will trim the sigma bounds and limit after the first one to make it easier to look at, but they are still present. $$Volume=\lim_{n\to\infty}\Sigma^{n}_{i=0}(\frac{6}{n}*\pi(144-16y^2))$$ $$Volume=\Sigma\frac{6\pi}{n}(144-16y^2)$$ $$Volume=\Sigma(\frac{6\pi}{n}*144)-\Sigma(\frac{6\pi}{n}*16y^2)$$ $$Volume=\Sigma(\frac{864\pi}{n})-\Sigma(\frac{96\pi y^2}{n})$$ $$Volume=864\pi-\frac{96\pi}{n}\Sigma y^2$$ $$Volume=864\pi-\frac{96\pi}{n}\Sigma(\frac{6i}{n}-3)^2$$ $$Volume=864\pi-\frac{96\pi}{n}\Sigma(\frac{36i^2}{n^2}-\frac{36i}{n}+9)$$ $$Volume=864\pi+\frac{96\pi}{n}\Sigma(-\frac{36i^2}{n^2}+\frac{36i}{n}-9)$$ $$Volume=864\pi-\frac{96\pi*36}{n^3}\Sigma i^2+\frac{96\pi*36}{n^2}\Sigma i-96\pi*9$$ These two identities are only true as $n$ approaches infinity: $$\frac{\Sigma i}{n^2}=\frac{1}{2}$$ $$\frac{\Sigma i^2}{n^3}=\frac{1}{3}$$ $$Volume=864\pi-\frac{96\pi*36}{3}+\frac{96\pi*36}{2}-864\pi$$ $$Volume=\frac{96\pi*36}{2}-\frac{96\pi*36}{3}$$ $$Volume=576\pi$$ $$\frac{576\pi}{4}=144\pi$$ Now we can set $144\pi$ equal to the sigma and solve for the depth. The $y$ bounds need to have a depth variable included. To find the volume $y$ was: $$y=\frac{6i}{n}-3$$ Notice that $6$ represents the depth from $y=-3$ (the bottom) which is what we want. Let $d$ represent depth. $$y=\frac{di}{n}-3$$ $$y^2=\frac{d^2i^2}{n^2}-\frac{6di}{n}+9$$ Changing the range of the interval to $d$ will change the $dy$ equivalent to $\frac{d}{n}$. We can steal this equation we came up with while solving volume as long as we account for the changing $dy$: $$Volume=864\pi-\frac{96\pi}{n}\Sigma y^2$$ $$144\pi=144d\pi-\frac{16d\pi}{n}\Sigma y^2$$ $$144=144d-\frac{16d}{n}\Sigma(\frac{d^2i^2}{n^2}-\frac{6di}{n}+9)$$ $$144=144d+\frac{16d}{n}\Sigma(-\frac{d^2i^2}{n^2}+\frac{6di}{n}-9)$$ $$144=144d-\frac{16d^3}{n^3}\Sigma i^2+\frac{16*6d^2}{n^2}\Sigma i-16d*9$$ $$144=144d-\frac{16d^3}{3}+\frac{96d^2}{2}-144d$$ $$144=\frac{96d^2}{2}-\frac{16d^3}{3}$$ $$144=48d^2-\frac{16d^3}{3}$$ $$432=144d^2-16d^3$$ $$27=9d^2-d^3$$ $$d^3-9d^2=-27$$ The value of $d$ can be approximated. $$d^2(d-9)=-27$$ $$d^2=\frac{-27}{d-9}$$ $$d=\pm\sqrt{\frac{27}{9-d}}$$ Start with a number that does not yield a negative number under the square root and then plug it back into itself many times until the number stops changing to enough decimal places. You can rearrange the equation various ways until the desired approximation is acquired.
Arrangements: $$d=\sqrt{\frac{27}{9-d}}$$ Yields $1.958110934$ $$d=-\sqrt{\frac{27}{9-d}}$$ Yields $-1.596266659$
$$d=9-\frac{27}{d^2}$$ Yields $8.638155725$

Only $1.958110934$ is between $0$ and $6$.
The depth is $1.958$ meters.

The following is the solution to solve it horizontally.

Solve for $x$ in terms of $y$ to get $x=\sqrt{144-16y^2}$. This is the right half. We can use method of rings to solve this problem, where we integrate from $y=0$ to $3$ and we use $\sqrt{144-16y^2}$ as our radius: $$\int_{-3}^3 \pi\left(\sqrt{144-16y^2}\right)^2dy=\int_{-3}^3 \pi\left(144-16y^2\right)dy=576\pi$$ Now, one-quarter of this is $144\pi$, so we want to find the depth when there is $144\pi$ fuel.

The round end of the ellipse is at $x=12$, so I'm going to assume we are filling from $x=12$ and then going to the left as we pour more propane in. Let's say we stop at $x=x_0$. If you try to visualize the shape of the propane tank, in the middle (vertically), it has a circle of radius $12$ and then it's built up of a bunch of ellipses horizontally. Therefore, we can draw a point from the center to the outer edge of the circle and then get a point at some $x$. Then, the major radius of that ellipse will be $\sqrt{12^2-x^2}=\sqrt{144-x^2}$ because the major radius of the ellipse and $x$ form a right triangle with hypotenuse the radius of the circle. Then, I assume that this ellipse is similar to our big ellipse, so we can say that the minor radius will be $\frac{\sqrt{144-x^2}}{4}$ since the minor radius in the original ellipse was one-fourth that of the major radius. Thus, we take the integral of the ellipse with these radii from $x=x_0$ to $12$: $$\int_{x_0}^{12} \pi(\sqrt{144-x^2})\left(\frac{\sqrt{144-x^2}}{4}\right)dx=\int_{x_0}^{12} \frac{\pi}{4}(144-x^2)dx \\ =-\frac{\pi}{4}\left(\frac{12^3}{3}-144*12\right)-\left(-\frac{\pi}{4}\left(\frac{x_0^3}{3}-144x_0\right)\right)=\frac{\pi}{4}\left(\frac{x_0^3}{3}-144x_0+1152\right)$$ Now, remember we want this to be $144\pi$, so we have: $$\frac{\pi}{4}\left(\frac{x_0^3}{3}-144x_0+1152\right)=144\pi$$ At this point, we need to use a computer algebra system to solve the cubic, at which point we get three different answers. However, only one of them is between $x=0$ and $x=12$, so we choose that one, giving us an answer of $x\approx 4.1676$.

Remember, though, we wanted to find the depth of the tank, which would be the distance from $x=12$ to $x\approx 4.1676$, meaning our answer is really $12-4.1676=7.8324$.

Now, this answer could be wrong because there's a lot of room to make mistakes here, so this might be wrong, but this is my best try at this problem. Good luck!

• Thank you. This isn't the correct answer, but I really appreciate you trying! – Rupali Jun 30 '16 at 23:11
• @Rupali Thanks for showing this problem to me! This was definitely a really interesting problem and I've never seen anything like this before, so it was a kind an interesting experience trying to solve it. – Noble Mushtak Jun 30 '16 at 23:13
• I think the mistake is in the interpretation of "depth". My best guess would be that actually the depth is along the $z$ axis instead (so that we have a tank which looks sort of like a sphere except that the $z=c$ cross sections are ellipses instead of circles). – Ian Jun 30 '16 at 23:23
• @Ian The cross sections across the $x$-direction is already ellipses, so I'm confused as to what you really mean, although it's certainly possible that I misinterpreted the question. At this point, I've been thoroughly confused as to how to solve this problem and I think I need a break for a while, so it will probably be best if you write the next answer if you know what to do. Thanks and good luck! – Noble Mushtak Jun 30 '16 at 23:31
• I think you want to set $\int_{-3}^{t}\pi(144-16y^2)dy=\frac{1}{4}(576\pi)$ and solve for $t$, and then the depth will be $t+3$ – user84413 Jul 1 '16 at 17:24