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11.7 – Spheres

Key Terms

  • Pythagorean Theorem – The theorem that relates the side lengths of a right triangle.
    • The theorem states that the square of the hypotenuse equals the sum of the squares of the legs: a^{2}+b^{2}=c^{2}
  • Axis – The diameter of a sphere.
  • Diameter – A line segment that contains the center of the circle and has endpoints on the circle.
    • This term also refers to the length of this line segment; the diameter of a circle is twice the radius.
  • Sphere – A three-dimensional figure consisting of all points in space that are the same distance from a given point.

Review

Remember
  • Surface Area
    • BA = base area
    • LA = lateral area
    • p = perimeter of base
    • h = height of solid
    • r = radius
    • s = slant height
  • Volume
    • B = base area
    • h = height
  • Surface area measures the total area of all the faces of a solid.
  • Volume measures the total amount of space inside a solid.

Notes

What is a Sphere?
  • How are circles and spheres alike?
    • They both consist of all the points that lie the same distance from a center point.
  • How are circles and spheres different?
    • Circles are two-dimensional, but spheres are three-dimensional.
  • Like a circle, a sphere has a radius and a diameter (called an axis). A sphere has an infinite number of radii and diameters.
    • The radius of a sphere is the distance from the center of the sphere to any point on the sphere.
  •  Diagram of a Sphere

GeoB 11.7 Sphere Labels

  • Surface Area of a Sphere

GeoB 11.7 SA of Spheres

  • Volume of a Sphere

GeoB 11.7 V of Spheres

  • Compare Surface Area & Volume of a Sphere
    • Both formulas use \pi
      • This makes sense because all cross-sections of spheres are circles.
    • Both formulas are multiplied by a constant.
      • You’ve seen this happen a lot with other solids. In fact, other volume formulas had a fraction constant with 3 on the bottom.
    • Both formulas include radius r.
      • This makes sense because a sphere is defined by its radius.
    • The surface area formula uses r^{2}.
      • You can remember the exponent of 2 for this formula because surface area answers are always given as units^{2}.
    • The volume formula uses r^{3}.
      • You can remember the exponent of 3 for this formula because volume answers are always given as units^{3}.

 

Surface Areas of Cylinders, Cones, and Spheres
  • The SA of a cone is less than the SA of a sphere, which is less than the SA of a cylinder.

GeoB 11.7 Compare r and d

GeoB 11.7 SA Solids Less ThanGeoB 11.7 Sphere Cylinder Cone

 

Volumes of Cylinders, Cones, and Spheres
  •  The volume of a cone is less than the volume of a sphere, which is less than the volume of a cylinder.

GeoB 11.7 V Solids Less ThanGeoB 11.7 V Comparisons

 

Surface Area of a Sphere
  • How to Find the Surface Area of Any Sphere
    • Step 1: Write the surface area formula.
    • Step 2: Find the radius of the sphere.
    • Step 3: Square the radius.
    • Step 4: Multiply the result by 4\pi.
  •  Example

GeoB 11.7 SA Sphere

  • SA (Sphere) = LA (Cylinder)

GeoB 11.7 SA Sphere LA Cylinder

 

Volume of a Sphere
  • How to Find the Volume of Any Sphere
    • Step 1: Write the volume formula.
    • Step 2: Find the radius of the sphere.
    • Step 3: Cube the radius.
    • Step 4: Multiply the result by \frac{4}{3}\pi.
  •  Example

GeoB 11.7 V of Sphere Ex

 

Pythagorean and Spheres
  • The cross-sectional areas of the cylinder and the sphere are found using the same general formula.
  • The Pythagorean theorem can be used to find the radius, a, of any cross-section of the sphere using b and r.
    • a^{2}+b^{2}=r^{2}
      • To isolate “a” (the radius), subtract b^{2} from both sides.
      • a^{2}=r^{2}-b^{2}
GeoB 11.7 Pythag

 

Volume of the Dugout Cylinder
  • Try to imagine removing (digging out) the two cones from the inside of the cylinder.
    • What’s left over is the dugout cylinder!
      • \pi r^{2} is the cross-sectional area of a cylinder, and \pi b^{2} is the cross-sectional area of an hourglass, or double-cone.
      • If you then subtract \pi b^{2} (the double-cone) from \pi r^{2} (the cylinder), you can make a new shape, called a dugout cylinder, whose cross-sectional area equals that of the sphere: \pi r^{2}-\pi b^{2}
  • To find the volume of a dugout cylinder:
    • Step 1: Find the volume of each, the cylinder and double-cone.
      • For this cylinder: BA=\pi r^{2} and h=2r, so V=Bh is \pi r^{2}\bullet 2r=2\pi r^{3}
      • For this double cone, multiply the volume of a cone by 2: V=2(\frac{1}{3}\pi r^{2}\bullet r)=\frac{2}{3}\pi r^{3}
        • Note, the height of EACH cone is r, not 2r.

GeoB 11.7 Dugout Cyl Cones

    • Step 2: Subtract the volume of the double-cone from the cylinder.

GeoB 11.7 Dugout Cyl Cones 2

Notice that the cylinder has the formula 2\pi r^{3}.
We can turn the 2 into a fraction, rewritten as \frac{6}{3}; and, \frac{6}{3}-\frac{2}{3}=\frac{4}{3},
which is why the formula for a dugout cylinder is \frac{4}{3}\pi r^{3}

 

Volume of the Dugout Cylinder and the Sphere
  • The Pythagorean theorem proves a sphere and dugout cylinder with equal heights have equal cross-sectional areas.
  • Cavalieri’s principle proves the sphere and dugout cylinder have equal volumes.

  • Volume of a Dugout Cylinder and a Sphere
    • Remember: a cross-section IS a circle for both solids below.
    • The Pythagorean theorem is a^{2}+b^{2}=c^{2}; and so, r^{2}+r^{2}=c^{2} because a = r and b = r for both!
    • The area of EACH cross section is \pi r^{2}, which can be called \pi a^{2} or \pi b^{2} since they the same thing.
      • We’ll use \pi a^{2} to represent the sphere and \pi b^{2} to represent the dugout part of the cylinder just for easy labeling of each solid.

GeoB 11.7 Dugout Cyl Sph Formula


  • Sphere + Double Cone = Cylinder

GeoB 11.7 Dugout Cyl Sph Formulas

\frac{4}{3}+\frac{2}{3}=\frac{6}{3}, and \frac{6}{3} simplifies to 2.


GeoB 11.7 Dugout Cylinder Sphere Volume

GeoB 11.7 Dugout Cyl and Sphere GeoB 11.7 Dugout Cylinder Sphere

Examples

  • Ex 1. What is the surface area of the sphere below?

GeoB 11.7 Ex01 Answer: 100\pi units^{2}

  • Ex 2. The axis of the sphere below is 16 units in length. What is the surface area of the sphere?

GeoB 11.7 Ex02

Answer: 256\pi units^{2}

  • Ex 3. What is the volume of the sphere below?

GeoB 11.7 Ex06

Answer: \frac{500}{3}\pi units^{3}

  • Ex 4. Given a sphere with radius r, the formula 4\pi r^{2} gives the surface area.
  • Ex 5. Which expression gives the volume of a sphere with radius 5?
    • Answer: \frac{4}{3}\pi 5^{3}
  • Ex 6. It is true that the area of a circle of radius 12 units is equal to the surface area of a sphere of radius 6 units.

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