Question

Given r=1 unit, find the volume of sphere?​

Answer 1

Answer:

The volume of the sphere with radius 1 unit is $4.19unit^{3}$.

Step-by-step explanation:

It is given that radius of a sphere is 1 unit.

A sphere is symmetrical, round in shape. It is a three dimensional solid, that has all its surface points at equal distances from the center. It has surface area and volume based on its radius.

It is asked to calculate the volume of the sphere.

We know that volume of the sphere$=\frac{4}{3}\pi r^{3}$

Value of $\pi =\frac{22}{7}$

Volume $=\frac{4}{3}*\frac{22}{7}*1^{3}$

            $=\frac{4}{3}*\frac{22}{7}$

            $=\frac{88}{21}=4.19$      

Volume of the sphere with radius 1 unit is $4.19unit^{3}$.

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Answer 2

Answer:

The volume of the sphere, when the radius is 1 unit= $\frac{4}{3}\pi$ units ==$\frac{88}{21}$=4.19 units

Step-by-step explanation:

To find the volume of a sphere with radius r=1 unit, we can use the formula:

V = (4/3)πr^3

where π is the mathematical constant pi, approximately equal to 3.14159.

Plugging in the value of r=1, we get:

V = (4/3)π(1)^3

V = (4/3)π

So the volume of the sphere is (4/3)π cubic units.

This for mula for the volume of a sphere can be derived using calculus. We can start by considering a sphere of radius r, divided into small slices with thick ness Δr. Each slice can be appro ximated as a cylind rical shell with height Δr and a circular base with radius r. The volume of each shell can be found using the formula for the volume of a cylinder, V = πr^2h.

The total volume of the sphere can then be found by adding up the volumes of all the shells as Δr approaches zero, resulting in an integral:

V = ∫0r 2πrh dr

V = ∫0r 2πr(r) dr

V = (2/3)πr^3

Simplifying this expression gives us the familiar formula for the volume of a sphere, V = (4/3)πr^3.

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