Volume of Composite Solids | Grade 8 Math | LevelKite

A composite solid is a solid made of simpler solids joined together, or a solid with a piece cut out of it. A grain silo is a cylinder with a dome on top, and an ice cream cone is a cone with a scoop on top. The dome and the scoop are each a hemisphere: half of a sphere, cut through its center, with a flat circular face.

Volume is the amount of space a solid fills, counted in cubic units. In each solid in this lesson, the two parts share one flat circle, the shared circle, so both parts have its radius. Joined parts meet only along that circle, and a flat circle has no thickness, so they do not overlap. That means you can find the space each part fills and add the two amounts. When one part is a piece cut out, its space is empty, so you take its volume away instead.

Key concept

Split, find each part, then add or subtract

Split the solid into named cylinders, cones and hemispheres that share one circle and one radius. Find each part's volume $V$ with the measures that part actually has: the shared radius $r$ and the part's own height, measured at a right angle to the shared circle. A cone's height runs from that circle straight to its tip; it is not the slant height, the length along its sloping side. A hemisphere's height is its radius, so it needs only $r$ .

Then add the volumes of joined parts, or subtract the volume of a part that is cut out. Each length is positive, and a circle's diameter (its width through the center) is $d = 2r$ . Use one unit for every length, so each volume is in that unit cubed. Keep $\pi$ as a symbol so the answer is exact.

V = V_{\text{part 1}} + V_{\text{part 2}} \quad \text{(joined)} \qquad V = V_{\text{uncut solid}} - V_{\text{cut-out part}} \quad \text{(cut out)}

Each part's volume, from the stated lengths in one unit

Part	Volume	Measures it uses
Cylinder	$\pi r^2 h$	Radius $r$ and its height $h$ ; $\pi r^2$ is the area of its circle, in square units
Cone	$\frac{1}{3}\pi r^2 h$	Radius $r$ and its height $h$ from circle to tip, not the slant height
Hemisphere	$\frac{1}{2} \cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3$	Radius $r$ only: it is half of a sphere with radius $r$ , so it has half the volume

The silo in the next example: a cylinder

10

m tall under a hemisphere dome, sharing a circle of radius

3

m, so the dome rises

3

m above the cylinder.

Worked example

A silo: cylinder plus hemisphere

A grain silo is a cylinder with a hemisphere dome on top. The shared circle has radius $3$ m, and the cylinder is $10$ m tall. Find the volume of the silo in terms of $\pi$ .

Name the parts and the measures each one has. The cylinder has radius $3$ m and height $10$ m. The dome is a hemisphere with the same $3$ m radius. All lengths are in meters, so the volumes will be in cubic meters.
$r = 3 \text{ m}, \quad h_{\text{cylinder}} = 10 \text{ m}$
Find the cylinder's volume with $\pi r^2 h$ . Its circle has area $\pi(3^2) = 9\pi$ square meters, and multiplying by the $10$ m height gives cubic meters.
$V_{\text{cylinder}} = \pi (3^2)(10) = \pi (9)(10) = 90\pi \text{ m}^3$
Find the dome's volume with $\frac{2}{3}\pi r^3$ , which needs only the radius. Cube it, $3^3 = 27$ , then take two thirds: $27 \div 3 = 9$ and $9 \times 2 = 18$ .
$V_{\text{hemisphere}} = \frac{2}{3}\pi (3^3) = \frac{2}{3}\pi (27) = 18\pi \text{ m}^3$
Add, because the dome is joined to the cylinder along the shared circle, so no space is counted twice. Both volumes are multiples of $\pi$ , so add the numbers in front: $90 + 18 = 108$ .
$V = 90\pi + 18\pi = 108\pi \text{ m}^3$

Answer

The silo's volume is exactly $108\pi$ cubic meters.

Worked example

An ice cream cone: cone plus hemisphere

An ice cream cone is packed full and topped with a hemisphere scoop. The scoop and the cone share a circle $6$ cm across. Measured straight up, the treat is $15$ cm tall from the tip of the cone to the top of the scoop. Find the volume of the ice cream in terms of $\pi$ .

Find the radius. The $6$ cm is the diameter of the shared circle, and $d = 2r$ , so divide it by $2$ .
$r = \frac{6}{2} = 3 \text{ cm}$
Find the cone's own height. The $15$ cm includes the scoop, which rises $3$ cm, its radius, above the shared circle, so subtract $3$ . What is left runs from the tip to the shared circle at a right angle to it, so it is the cone's height, not a slant height.
$h_{\text{cone}} = 15 - 3 = 12 \text{ cm}$
Find the cone's volume with $\frac{1}{3}\pi r^2 h$ , using its own $12$ cm height.
$V_{\text{cone}} = \frac{1}{3}\pi (3^2)(12) = \frac{1}{3}\pi (9)(12) = \frac{1}{3}\pi (108) = 36\pi \text{ cm}^3$
Find the scoop's volume with $\frac{2}{3}\pi r^3$ . The arithmetic matches the silo's dome, but the unit is now centimeters.
$V_{\text{hemisphere}} = \frac{2}{3}\pi (3^3) = \frac{2}{3}\pi (27) = 18\pi \text{ cm}^3$
Add, because the scoop is joined to the cone along the shared circle.
$V = 36\pi + 18\pi = 54\pi \text{ cm}^3$

Answer

The treat holds exactly $54\pi$ cubic centimeters of ice cream.

The treat from the example above: a cone

12

cm tall and a scoop sharing a circle of radius

3

cm, the measures from steps 1 and 2, with the scoop's

3

cm on top making

15

cm in all.

Some composite solids have a part cut out. In the figure below, a cone-shaped hollow, an empty space, goes down into a solid cylinder from its top circle. The solid that remains is the cylinder without the hollow's space, so its volume is the cylinder's volume minus the hollow's volume.

This works only if the whole hollow lies inside the cylinder, and two facts make sure it does. The hollow shares the cylinder's top circle, so it is exactly as wide. Its depth, measured at a right angle to that circle, is no more than the cylinder's height. So the hollow fills only part of the cylinder's space, and the difference is positive.

The solid in the next example: a cylinder

9

cm tall with radius

4

cm (half its

8

cm width) and a cone hollow

6

cm deep, which fits inside because

6 \le 9

.

Worked example

A cylinder with a cone cut out

A solid cylinder is $8$ cm across and $9$ cm tall. A cone-shaped hollow, $6$ cm deep, goes down into it from its top circle and shares that circle. Find the volume of the solid that remains, in terms of $\pi$ .

Find the radius, half of the $8$ cm diameter, since $d = 2r$ . Then check the fit: the hollow is $6$ cm deep and the cylinder is $9$ cm tall, so the hollow lies inside. All lengths are in centimeters.
$r = \frac{8}{2} = 4 \text{ cm}, \quad 6 \text{ cm} \le 9 \text{ cm}$
Find the volume of the whole cylinder as if it had no hollow. This is the space the hollow comes out of.
$V_{\text{cylinder}} = \pi (4^2)(9) = \pi (16)(9) = 144\pi \text{ cm}^3$
Find the hollow's volume. It is a cone with the shared $4$ cm radius, and its height is its $6$ cm depth.
$V_{\text{cone}} = \frac{1}{3}\pi (4^2)(6) = \frac{1}{3}\pi (16)(6) = \frac{1}{3}\pi (96) = 32\pi \text{ cm}^3$
Subtract, because the hollow's space is missing from the cylinder: $144 - 32 = 112$ . The result is positive, as it must be for a hollow that fits inside.
$V = 144\pi - 32\pi = 112\pi \text{ cm}^3$

Answer

The solid that remains has a volume of exactly $112\pi$ cubic centimeters.

Takeaway: Split a composite solid into named cylinders, cones and hemispheres that share one circle and one radius. Find each part's volume with the measures it actually has: the shared radius and its own height, measured at a right angle to that circle, with every length in one unit. Add the volumes of joined parts, subtract the volume of a part that is cut out, and keep $\pi$ so the answer stays exact. Next, you will compare the volumes of cylinders, cones and spheres.