3. An expandable cone-shaped funnel consists of two sections as shown. (a) What is the volume of the bottom section? (b) What is the volume of the top section?

1. Volume of cone = [tex] \frac{1}{3}h \pi r^{2} [/tex]

where h is the height of the cone and r is the radius.

2. The bottom section is a cone with radius 6/2 = 3 (cm) and height= 8 cm

so the volume = [tex]\frac{1}{3}h \pi r^{2} =\frac{1}{3}*8 *\pi *3^{2}=24 \pi [/tex] ([tex] cm^{3} [/tex])

3. The volume of the top section is the volume of the cone - volume of bottom cone, which we already found.

V(cone)=[tex]\frac{1}{3}h \pi r^{2}=\frac{1}{3}*16* \pi* 6^{2}=192 \pi [/tex]

V(top section)=192[tex] \pi [/tex]-24[tex] \pi [/tex]=168[tex] \pi [/tex] [tex]( cm^{3} )[/tex]

AFOKE88 AFOKE88 · Answer 2 · 2022-07-07T12:02:52+02:00

A) Volume of bottom of cone = 24π cm³

B) Volume of top section = 168π cm³

How to find the Volume of a Cone?

Formula for volume of a cone is;

V = ¹/₃πr²h

A) The bottom section is a cone that has;

radius; r = 6/2 = 3 cm

height; h = 8 cm

Thus;

V = ¹/₃π(3)²*8

V = 24π cm³

b) Volume of cone is;

V = ¹/₃π(6)²*16

V = 192π cm³

Thus;

Volume of the top section is;

V = 192π - 24π

V = 168π cm³

Read more about Cone Volume at; https://brainly.com/question/12004994

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