Cylinder Surface Area = 2 π r h + 2 π r 2 , Volume = π r 2 h \text{Surface Area} = 2\pi rh + 2\pi r^2, \qquad \text{Volume} = \pi r^2 h Surface Area = 2 π r h + 2 π r 2 , Volume = π r 2 h Worked example: find the surface area and volume of a cylinder with r = 7 r=7 r = 7 h = 10 h=10 h = 10 π ≈ 22 7 \pi \approx \tfrac{22}{7} π ≈ 7 22
Volume = 22 7 × 49 × 10 = 22 × 7 × 10 = 1540 \text{Volume} = \dfrac{22}{7} \times 49 \times 10 = 22 \times 7 \times 10 = 1540 Volume = 7 22 × 49 × 10 = 22 × 7 × 10 = 1540 Surface Area = 2 × 22 7 × 7 × ( 10 + 7 ) = 44 × 17 = 748 \text{Surface Area} = 2 \times \dfrac{22}{7} \times 7 \times (10+7) = 44 \times 17 = 748 Surface Area = 2 × 7 22 × 7 × ( 10 + 7 ) = 44 × 17 = 748 Cone and its frustum A cone's slant height is l = r 2 + h 2 l = \sqrt{r^2+h^2} l = r 2 + h 2
Surface Area = π r ( r + h 2 + r 2 ) , Volume = 1 3 π r 2 h \text{Surface Area} = \pi r (r + \sqrt{h^2+r^2}), \qquad \text{Volume} = \dfrac{1}{3}\pi r^2 h Surface Area = π r ( r + h 2 + r 2 ) , Volume = 3 1 π r 2 h Slicing the top off a cone parallel to its base leaves a frustum , with base radius R R R r r r h h h R − r R-r R − r
s = ( R − r ) 2 + h 2 s = \sqrt{(R-r)^2 + h^2} s = ( R − r ) 2 + h 2 Worked example: a frustum has R = 5 R=5 R = 5 r = 2 r=2 r = 2 h = 4 h=4 h = 4
s = ( 5 − 2 ) 2 + 4 2 = 9 + 16 = 25 = 5 s = \sqrt{(5-2)^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 s = ( 5 − 2 ) 2 + 4 2 = 9 + 16 = 25 = 5 The lateral surface area then follows directly:
Lateral Surface Area = π ( R + r ) s = π ( 5 + 2 ) ( 5 ) = 35 π \text{Lateral Surface Area} = \pi(R+r)s = \pi(5+2)(5) = 35\pi Lateral Surface Area = π ( R + r ) s = π ( 5 + 2 ) ( 5 ) = 35 π