Let 𝑉 be the volume of a right circular cone of height ℎ=28 whose base is a circle of radius 𝑅=4.

Question

(a) Use similar triangles to find the area of a horizontal cross section at a height 𝑦. Give your answer in terms of 𝑦.

(b) Calculate 𝑉by integrating the cross‑sectional area.

Answer 1

V = hB/3 = hpi(R^2)/3 B = pi(R^2) = Area of the Base

if h=28 , then V = 28piR^2/3

and if R=4, then V =28pi(4)^2/3 = 16(28)pi/3

pi = about 3.14

V= 16(28)(3.14)/3 = about 469.14 cubic units for volume of the entire cone

a) volume of the lower part of the cone sliced horizontally off the top at height = y is:

v = hB/3 - (h-y)b/3 = (B-yb)/3

where B is the Base of the cone and

b is the base of a horizontal slice of the cone at height y.

b is the area of the smaller base = pi(r^2) where r= the radius of the smaller circle

a) construct a right triangle within the cone of height 28, base 4= radius, hypotenuse = slant height

then construct a smaller right triangle with height h-y = 28-y and base = small r

28/4 = 7 = (28-y)/r ratios of height & base = constant

7r = 28-y

r = (28-y)/7 = 4 -y/7

0<y<28

0<r<4

area of the cross section = pir^2 = pi(4-y/7)^2

area = pi(16 -8y/7 + y^2/49)

= about (3.14)(16) - (3.14)8y/7 + (3.14)y^2/49

b) integrate (50.24 - 3.59y + .064y^2) dy

= 50.24y -3.59y^2/2 + .064y^3/3 evaluated from 0 to y

if y = 0 volume = 0

if y= 28, volume = 50.24(28) -3.59(28^2)/2 + .064(28)^3/3 = 1406.7 - 1407.28 + 468.9

= 468.3 which compares to 469.1 calculated earlier, close enough to be explained by rounding errors, just 0.8 difference