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Surface Areas and Volumes - Volume of a Sphere and Hemisphere

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A sphere is a perfectly round three-dimensional solid figure where every point on the surface is at an equal distance, known as the radius rr, from a fixed central point. Visually, it looks like a completely round ball.

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The volume of a sphere represents the total capacity or space occupied by it. If you were to fill a hollow sphere with water, the amount of water it holds is its volume, calculated using the formula 43πr3\frac{4}{3} \pi r^3.

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A hemisphere is exactly one-half of a sphere, created when a sphere is cut by a plane passing through its center. Visually, it resembles a bowl with a flat circular top and a curved bottom.

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The volume of a hemisphere is exactly half the volume of a sphere of the same radius. Therefore, the formula used is 23πr3\frac{2}{3} \pi r^3.

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The radius rr is the distance from the center to any point on the boundary. If the diameter dd (the widest distance across the sphere) is provided, it must be halved to find the radius using the relation r=d2r = \frac{d}{2}.

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Volume is always expressed in cubic units. Common units include cubic centimeters (cm3cm^3), cubic meters (m3m^3), or liters (LL). Note that 1000cm3=11000 cm^3 = 1 liter.

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In real-world applications, such as finding the weight of a metallic shot-put, the mass can be calculated if the volume and density are known, using the formula: Mass=Volume×DensityMass = Volume \times Density.

📐Formulae

Volume of a Sphere=43πr3\text{Volume of a Sphere} = \frac{4}{3} \pi r^3

Volume of a Hemisphere=23πr3\text{Volume of a Hemisphere} = \frac{2}{3} \pi r^3

Radius (r)=d2 (where d is the diameter)\text{Radius (r)} = \frac{d}{2} \text{ (where d is the diameter)}

π≈227 or 3.14\pi \approx \frac{22}{7} \text{ or } 3.14

💡Examples

Problem 1:

Find the volume of a sphere whose radius is 10.5cm10.5 cm. (Take π=227\pi = \frac{22}{7})

Solution:

  1. Given: Radius r=10.5cm=212cmr = 10.5 cm = \frac{21}{2} cm.
  2. Formula for volume of a sphere: V=43πr3V = \frac{4}{3} \pi r^3.
  3. Substitute the values: V=43×227×(212)3V = \frac{4}{3} \times \frac{22}{7} \times (\frac{21}{2})^3.
  4. Expand: V=43×227×212×212×212V = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}.
  5. Simplify by cancelling: V=11×21×212=48511=4851cm3V = 11 \times 21 \times \frac{21}{2} = \frac{4851}{1} = 4851 cm^3.

Explanation:

To find the sphere's volume, we plug the radius into the standard formula. Converting the decimal 10.510.5 to the fraction 212\frac{21}{2} simplifies the calculation by allowing easy cancellation with the numbers 33 and 77 in the denominator.

Problem 2:

A hemispherical tank is made up of an iron sheet 1cm1 cm thick. If the inner radius is 1m1 m, then find the volume of the iron used to make the tank.

Solution:

  1. Inner radius (rr) = 1m=100cm1 m = 100 cm.
  2. Thickness = 1cm1 cm.
  3. Outer radius (RR) = Innerradius+Thickness=100+1=101cmInner radius + Thickness = 100 + 1 = 101 cm.
  4. Volume of iron used = External Volume - Internal Volume.
  5. V=23πR3−23πr3=23π(R3−r3)V = \frac{2}{3} \pi R^3 - \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (R^3 - r^3).
  6. V=23×227×(1013−1003)V = \frac{2}{3} \times \frac{22}{7} \times (101^3 - 100^3).
  7. V=4421×(1030301−1000000)=4421×30301V = \frac{44}{21} \times (1030301 - 1000000) = \frac{44}{21} \times 30301.
  8. V≈63487.81cm3V \approx 63487.81 cm^3 or 0.06348m30.06348 m^3.

Explanation:

The volume of the material used in a hollow object is the difference between the outer volume and the inner volume. We first calculate the outer radius by adding thickness to the inner radius, then apply the hemisphere volume formula to both and subtract.