Question

3. A balancing balloon toy is in the shape of a hemisphere (half-sphere) attached to the base of a cone. If the toy is 4ft tall and 2ft wide, what is the volume of the toy? Round your answer to two decimal places.

Answer 1

Answer:

The volume of the toy is $V=5.23\ ft^3$

Step-by-step explanation:

step 1

Find the volume of the hemisphere

The volume of the hemisphere is given by the formula

$V=\frac{2}{3}\pi r^{3}$

In this problem, the wide of the toy is equal to the diameter of the hemisphere

so

$D=2\ ft$

$r=2/2=1\ ft$ ----> the radius is half the diameter

substitute

$V=\frac{2}{3} \pi (1)^{3}=\frac{2}{3} \pi\ ft^3$

step 2

Find the volume of the cone

The volume of the cone is given by

$V=\frac{1}{3}\pi r^{2}h$

we know that

The radius of the cone is the same that the radius of the hemisphere

so

$r=1\ ft$

The height of the cone is equal to subtract the radius of the hemisphere from the height of the toy

$h=4-1=3\ ft$

substitute the given values

$V=\frac{1}{3}\pi (1)^{2}(3)=\pi\ ft^3$

step 3

Find the volume of the toy

we know that

The volume of the toy, is equal to the volume of the cone plus the volume of the hemisphere.

so

$V=(\frac{2}{3} \pi+\pi)\ ft^3$

$V=(\frac{5}{3}\pi)\ ft^3$

assume

$\pi=3.14$

$V=\frac{5}{3}(3.14)=5.23\ ft^3$