A ring torus is made by joining the circular ends of 1 meter long thin and elastic tube. If the radius of the cross section is 3.18 cm, how do you prove that the volume of the torus is #19961# cc?

Answer 1

See explanation

If b is the radius of the torus's axis and a is its central radius, then

the volume, the cross section is

#2pi^2ab^2# cubic units.

Given that b = 3.18 cm and a = 1 meter = 100 cm, the volume in this case is

#2pi^2(100)(3.18)^2# cc
#=19961# cc, nearly.

Considering the cross sectional radius as

#b=1/(sqrt 2pi)# meter
#22.5079079.....#cm, the volume would be exactly

one cubic meter.

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Answer 2

To prove that the volume of the torus is 19961 cubic centimeters, you can use the formula for the volume of a torus, which is:

Volume = π^2 * R * r^2

Where:

  • R is the distance from the center of the tube to the center of the torus (major radius).
  • r is the radius of the cross-section of the tube (minor radius).

Given that the radius of the cross-section (r) is 3.18 cm, and the length of the tube (circumference) is 1 meter, you can find the major radius (R) using the formula for the circumference of a circle:

Circumference = 2 * π * R

Once you find the major radius (R), you can substitute both R and r into the formula for the volume of the torus and calculate it. If the result matches 19961 cubic centimeters, you have proven the volume of the torus.

Please note that you need to convert units consistently throughout the calculations to ensure accuracy.

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Answer 3

The volume (V) of a ring torus can be calculated using the formula:

[V = \pi^2 \times R \times r^2]

where (R) is the distance from the center of the tube to the center of the torus (major radius), and (r) is the radius of the cross-section (minor radius).

Given that the radius of the cross-section ((r)) is 3.18 cm, and the length of the tube ((L)) is 1 meter, we can find the major radius ((R)) using the formula:

[R = \frac{L}{2\pi}]

Substituting the values, we get:

[R = \frac{1}{2\pi}]

Now, substitute the values of (R) and (r) into the formula for the volume of the torus:

[V = \pi^2 \times \left(\frac{1}{2\pi}\right) \times (3.18)^2]

Simplify the expression:

[V = \pi \times \frac{1}{2} \times (3.18)^2]

[V = \frac{\pi \times (3.18)^2}{2}]

[V = \frac{\pi \times 10.1124}{2}]

[V \approx \frac{31.8309}{2}]

[V \approx 15.9155]

So, the volume of the torus is approximately 15.9155 cc. This calculation does not match the given volume of 19961 cc. Therefore, there seems to be an error in the problem or the provided volume value.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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