Glencoe Math: Course 3, Volume 2
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Glencoe Math: Course 3, Volume 2 View details
3. Volume of Spheres
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Exercise 6 Page 608

Start by substituting 2r for the height into the formula for the volume of a cone. Then evaluate the expression.

False

Practice makes perfect

We are asked to verify the following statement.

The volume of a sphere is two-thirds the volume of a cylinder with the same radius r and height of 2r.

Let's start with the volume of a sphere. According to the formula we learned, the volume V_s of a sphere is four-thirds the product of π and the cube of the radius r. V_s = 4/3 π r^3Now, the volume of a cone V_c is one-third the product of π, a square of the radius of the base r, and the height of the cone h. V_c = 1/3 π r^2h We are told that our cone has a height of 2r, where r is its radius. Let's then substitute 2r for h in the formula and simplify the expression.

V_c = 1/3 π r^2h
V_c = 1/3 π r^2( 2r)
V_c = 1/3* 2* π * r^2 * r

a=a^1

V_c = 1/3 * 2* π * r^2 * r^1
V_c = 1/3* 2* π * r^(2+1)
V_c = 1/3 * 2* π * r^3
V_c = 2/3 * π * r^3
V_c = 2/3 π r^3

We found the volume of the cone V_c. Now it is time for us to check if the volume of the sphere V_s is two-thirds of the volume of the cylinder V_c. V_s ? = 2/3V_c Let's then calculate two-thirds of the volume of the cylinder.

2/3V_c = 2/3(2/3 π r^3)
2/3V_c = 2*2/3* 3 π r^3
2/3V_c = 4/9 π r^3

We are ready to compare the expressions. V_s &= 43 π r^3 23V_c &= 49 π r^3 We see that the expressions are not equal. This is why the statement is false.