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I am trying to calculate the area of a truncated cone, from its volume, height, and other area. On the image below, I know $h$, $V$, and $a_1$ (the area of the base of the truncated cone). I am not looking for the radius, just the area.

illustration of a truncated cone

My object is not a perfect truncated cone: I am trying to calculate the area of a lake at $10 m$ below the surface. I know the surface of the lake ($4.3 km^2$), the volume of the $0-10 m$ section ($0.03 km^3$), and $h = 0.010 km$. In other word, based on the drawing above, the equation I am trying to solve is (source here):

$V = \frac{1}3 \pi (r_1^2 + r_1 r_2 + r_2^2) h$

with,
$r_1 = \sqrt{\frac{4.3}\pi} \approx 1.17 $
$h = 0.01$
$V = 0.03$

$r_2$ is unknown.

I got as far as:

$0.03 = \frac{1}3 \pi (1.17^2 + 1.17 r_2 + r_2^2) 10$
$\frac{0.09}{10 \pi} = 1.17^2 + 1.17 r_2 + r_2^2 $
$\frac{0.09}{10 \pi} - 1.17^2 = 1.17 r_2 + r_2^2 $
$-1.366 = 1.17 r_2 + r_2^2 $

Basically, I have a formula such as $y = ax + x^2$, and I need to find $x$ knowing $a$ and $y$.

Any help would be greatly appreciated!



Log of changes:

  • A previous version of my question had a wrong volume: $0.3 km^3$ instead of $0.03 km^3$
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  • $\begingroup$ Are you looking for the total area of all sides, or just the top ellipse? $\endgroup$ – user20672 Apr 8 at 2:25
  • $\begingroup$ Using similarity you could calculate the information of the non truncated cone. $\endgroup$ – Moti Apr 8 at 6:47
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You have a mistake in $$0.03 = \frac{1}3 \pi (1.17^2 + 1.17 r_2 + r_2^2) 10$$

It should be $$0.03 = \frac{1}3 \pi (1.17^2 + 1.17 r_2 + r_2^2) 0.01$$

Thus you get (with rounding): $$ 2.86 = 1.17^2 + 1.17r_2 + r_2^2 \Leftrightarrow$$ $$r_2^2 + 1.17r_2 + (1.17^2 - 2.86) = 0 \Leftrightarrow$$ $$r_2^2 + 1.17r_2 -1.5 = 0$$

This is a quadratic equation (an equation of form $ax^2+bx+c=0$).

You can solve this using the formula for solving quadratic equations: $$x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}$$

You can get more information on the formula here or even solve it online.

In your case $a = 1$, $b = 1.17$ and $c = -1.5$. Plugging those values into the formula you will get two solutions: ~-1.94229 and ~0.772286.

Since $r_2 > 0$ only the second root is valid and also your solution.

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  • $\begingroup$ Great, good catch!! Thank you very much. If that doesn't take too long, could you write down the step by step conversion from my ax^2$ + bx +c = 0 to x ~ 0.7723, please? My calculus memories are failing me and I can't solve this anymore... I will need to repeat this for several truncated cone, so need to understand the steps. $\endgroup$ – Rosalie Bruel Apr 9 at 0:29
  • $\begingroup$ Found it! Thanks again for your help. montereyinstitute.org/courses/DevelopmentalMath/… $\endgroup$ – Rosalie Bruel Apr 9 at 15:09
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Welcome to MSE. First of all area of the lake can not be $4.3 km^2$ becase even if the lake is cylinder the volume will be $4.3\times 0.01=0.043 km^3$ not $0.3 km^3$. So I think the surface area must be $43 km^2$.

The cross section of the lake is trapezoid. We can consider a rectangle with width average of diameters of upper and bottom surfaces. In this case we consider a cylinder instead of a truncated cone.Let diameter of surface be $2r$ and that of bottom be $2r_1$ and corresponding diameter of the cylinder be $2r_a$, we have:

$$2r_a=\frac{2r+2r_1}2\rightarrow r_a=\frac{r+r_1}2$$

$$r=\sqrt{\frac {43}{3.14}}\approx 3.7$$

$$\big(\frac{r+r_1}2\big)^2\times 3.14\times 0.01=0.3\rightarrow r_1+r=6.8$$

Hence the radius of bottom area is:

$6.8-3.7=3.1$

And it area is:

$$A=3.1^2\times 3.14\approx 30.2 km^2$$

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  • $\begingroup$ Thanks for your answer, but unfortunately, this doesn't seem to be what I am looking for. Using your values of $r_1$ and $r_2$, I tried calculating the volume of my truncated cone, and I don't get the value I am looking for. I am going to edit my question to make it hopefully clearer. You were right though, I had given the wrong volume! I edited that in my question as well. $\endgroup$ – Rosalie Bruel Apr 8 at 20:48

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