# Calculate the top area of a truncated cone with known volume, height, and bottom area

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 $$0.03 km^3$$
• Are you looking for the total area of all sides, or just the top ellipse? – user20672 Apr 8 at 2:25
• Using similarity you could calculate the information of the non truncated cone. – Moti Apr 8 at 6:47

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.

• 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. – Rosalie Bruel Apr 9 at 0:29 • Found it! Thanks again for your help. montereyinstitute.org/courses/DevelopmentalMath/… – Rosalie Bruel Apr 9 at 15:09 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: $$A=3.1^2\times 3.14\approx 30.2 km^2$$ • 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. – Rosalie Bruel Apr 8 at 20:48