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 $0.03 km^3$
Best Answer
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:
$$A=3.1^2\times 3.14\approx 30.2 km^2$$