To calculate the volume of water in a conical tank, we start with the formula for the volume of a cone:
\[ V = \frac{1}{3} \pi r^2 h \]
Where:
- \( V \) is the volume,
- \( r \) is the radius of the base,
- \( h \) is the height.
In this problem, the tank is an inverted cone with a total height of 6 feet and a radius of 2 feet. We are interested in the volume of the water when the water level is at 3.5 feet.
First, we need to find the radius of the cone where the water level is 3.5 feet. Since the cone is similar at different heights, we can use the ratio of the heights and radii.
Given:
- Full height of the cone, \( H = 6 \) feet
- Full radius of the cone, \( R = 2 \) feet
- Water height, \( h = 3.5 \) feet
We can determine the radius \( r \) of the water surface at height \( h \) using the similarity of triangles:
\[ \frac{r}{R} = \frac{h}{H} \]
Substituting in our values:
\[ \frac{r}{2} = \frac{3.5}{6} \]
Now, solve for \( r \):
\[ r = 2 \cdot \frac{3.5}{6} = 2 \cdot \frac{3.5}{6} = \frac{7}{6} \text{ feet} \]
Now we know that the radius of the water surface is \( \frac{7}{6} \) feet at a height of 3.5 feet.
Next, we can calculate the volume of the water using the volume formula for a cone:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting in \( r = \frac{7}{6} \) feet and \( h = 3.5 \) feet:
\[ V = \frac{1}{3} \pi \left(\frac{7}{6}\right)^2 (3.5) \]
Calculating \( \left(\frac{7}{6}\right)^2 \):
\[ \left(\frac{7}{6}\right)^2 = \frac{49}{36} \]
Now substituting \( \frac{49}{36} \) back into the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{49}{36}\right) (3.5) \]
Convert \( 3.5 \) to a fraction:
\[ 3.5 = \frac{7}{2} \]
So we substitute that in:
\[ V = \frac{1}{3} \pi \left(\frac{49}{36}\right) \left(\frac{7}{2}\right) \]
Multiplying:
\[ = \frac{1}{3} \pi \cdot \frac{49 \cdot 7}{36 \cdot 2} \] \[ = \frac{1}{3} \pi \cdot \frac{343}{72} \]
Perform the multiplication with \( \frac{1}{3} \):
\[ = \frac{343 \pi}{216} \]
Thus, the volume of the water in the tank when the water level is 3.5 feet is:
\[ \boxed{\frac{343 \pi}{216}} \text{ cubic feet} \]
If we want to numerically approximate this volume (using \( \pi \approx 3.14159 \)):
\[ V \approx \frac{343 \cdot 3.14159}{216} \approx 5.0 \text{ cubic feet} \text{ (approximately)} \]
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