Let's analyze the given composite solid.

The solid consists of two smaller ones.

A cylinder with a radius of $r = 4$ inches, and the height of $h = 5$ inches.
A hemisphere with a radius of $r = 4$ inches.

We are asked to find the volume and surface area of the above solid. First, let's find the volume.

Volume

Let's use the formula for the volume of a sphere and for the volume of a cylinder.

Solid	Cylinder	Hemisphere
Radius	$r = 4$	$r = 4$
Height	$h = 5$	$-$
Volume	$V = π r^{2} h$	$V = \frac{1}{2} \cdot \frac{4}{3} π r^{3}$
Volume	$V_{1} = π (4)^{2} (5) = 80 π$	$V_{2} = \frac{1}{2} \cdot \frac{4}{3} π (4)^{3} = \frac{1 2 8}{3} π$

The volume of the cylinder is

80 π

cubic inches and the volume of the hemisphere is

\frac{1 2 8}{3} π

cubic inches. Now, let's add these volumes to get the volume of the given composite solid,

V_{solid} .

We will also round the answer to the nearest tenth.

V_{solid} = V_{1} + V_{2}

▼

Substitute values and evaluate

SubstituteII

$V_{1} = 80 π$ , $V_{2} = \frac{1 2 8}{3} π$

V_{solid} = 80 π + \frac{1 2 8}{3} π

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V_{solid} = 80 π + \frac{1 2 8 π}{3}

NumberToFrac

$a = \frac{3 \cdot a}{3}$

V_{solid} = \frac{3 \cdot 8 0 π}{3} + \frac{1 2 8 π}{3}

Multiply

V_{solid} = \frac{2 4 0 π}{3} + \frac{1 2 8 π}{3}

AddFrac

Add fractions

V_{solid} = \frac{3 6 8 π}{3}

UseCalc

Use a calculator

V_{solid} = 385.3686988 \dots

RoundDec

Round to $1$ decimal place(s)

V_{solid} \approx 385.4

We found that the volume of the composite solid is about

385.4

cubic inches.

Surface Area

Now,we will find the surface area of the given composite solid. Notice that it is equal to the surface area of the hemisphere, the lateral area of the cylinder, and one base area of the cylinder. Let's use the formulas for the surface area of a sphere and for the surface area of a cylinder.

Surface	Hempishere	Lateral Area of Cylinder	Base Area of Cylinder
Radius	$r = 4$	$r = 4$	$r = 4$
Height	$h = 5$	$-$	$-$
Area	$A = \frac{1}{2} \cdot 4 π r^{2}$	$A = 2 π r h$	$A = π r^{2}$
Area	$A_{1} = \frac{1}{2} \cdot 4 π (4)^{2} = 32 π$	$A_{2} = 2 π (4) (5) = 40 π$	$A_{3} = π (4)^{2} = 16 π$

Exercise