The volume of sphere is the amount of liquid a sphere can hold. It is the space occupied by a sphere in 3-dimensional space. It is measured in unit3 i.e. m3, cm3, etc. Show A sphere is a three-dimensional solid object with a round form in geometry. From a mathematical standpoint, it is a three-dimensional combination of a group of points connected by one common point at equal distances. A sphere, unlike other three-dimensional shapes, has no vertices or edges. Its centre is equidistant from all places on its surface. In other words, the distance between the sphere’s centre and any point on its surface is the same. Various spherical objects used in daily life are football, basketball, Earth, Moon, etc. The volume of a sphere is the amount of space it takes up within it. The sphere is a three-dimensional round solid shape in which all points on its surface are equally spaced from its centre. The fixed distance is known as the sphere’s radius, and the fixed point is known as the sphere’s centre. We will notice a change in form when the circle is turned. As a result of the rotation of the two-dimensional object known as a circle, the three-dimensional shape of a sphere is obtained. The formula for the volume of a sphere is given by,
Using the integration approach, we can simply calculate the volume of a sphere.
How to Calculate Volume of Sphere?Volume of sphere is the space occupied by a sphere. Its volume can be calculated using the formula V = 4/3πr3. Steps required to calculate the volume of a sphere are:
Example: Find the volume of a sphere with a radius of 7 cm. Solution:
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Solved Examples on Volume of SphereExample 1. Find the volume of the sphere whose radius is 9 cm. Solution:
Example 2. Find the volume of the sphere whose radius is 12 cm. Solution:
Example 3. Find the volume of the sphere whose radius is 6 cm. Solution:
Example 4. Find the volume of the sphere whose radius is 4 cm. Solution:
Example 5. Find the volume of the sphere whose diameter is 10 cm. Solution:
Example 6. Find the volume of the sphere whose diameter is 16 cm. Solution:
Example 7. Find the volume of the sphere whose diameter is 14 cm. Solution:
FAQs on Volume of SphereQuestion 1: Write the formula for the total surface area of the sphere. Answer:
Questions 2: What is the formula for the Volume of the Sphere? Answer:
Question 3: How do we find the volume of Hemi-Sphere? Answer:
Question 4: If a sphere and a hemisphere have the same radii then what is the ratio of their volume? Answer:
Question 5: How do we measure the volume of Sphere? Answer:
r = radius V = volume A = surface area C = circumference π = pi = 3.1415926535898 √ = square root Calculator UseThis online calculator will calculate the 3 unknown values of a sphere given any 1 known variable including radius r, surface area A, volume V and circumference C. It will also give the answers for volume, surface area and circumference in terms of PI π. A sphere is a set of points in three dimensional space that are located at an equal distance r (the radius) from a given point (the center point). Units: Note that units are shown for convenience but do not affect the calculations. The units are in place to give an indication of the order of the results such as ft, ft2 or ft3. For example, if you are starting with mm and you know r in mm, your calculations will result with A in mm2, V in mm3 and C in mm. Sphere Formulas in terms of radius r:
\[ V = \frac{4}{3}\pi r^3 \] \[ V \approx 4.1888r^3 \]
\[ V = \frac{A^{3/2}}{6\sqrt{\pi}} \] \[ V \approx 0.09403A^{3/2} \]
\[ V = \frac{C^3}{6\pi^2} \] \[ V \approx 0.01689C^3 \]
\[ A \approx 12.5664r^2 \]
\[ A = \pi^{1/3} (6V)^{2/3} \] \[ A \approx 4.83598V^{2/3} \]
\[ A = \frac{C^2}{\pi} \] \[ A \approx 0.3183C^2 \]
\[ r = \left(\frac{3V}{4 \pi}\right)^{1/3} \] \[ r \approx 0.62035V^{1/3} \]
\[ r = \sqrt{\frac{A}{4 \pi}} \] \[ r \approx 0.2821 \sqrt{A} \]
\[ r = \frac{C}{2 \pi} \]
\[ C = \pi^{2/3} (6V)^{1/3} \] \[ C \approx 3.89778V^{1/3} \]
\[ C \approx 1.77245\sqrt{A} \] Sphere Calculations:Use the following additional formulas along with the formulas above.
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