Question

1. filling the punch bowl

you will need to determine the number of times that the punch bowl will need to be filled during the party. the punch bowl you are using is a hemisphere (half - sphere) with a diameter of 35 cm. it can be filled to 85% capacity.
a. calculate the capacity of the punch bowl. round your answer to two decimal places.
hint: $v_{sphere}=\frac{4}{3}pi r^{3}$
b. determine the volume of the punch in the bowl if the bowl can be filled to 85% capacity. round your answer to the nearest $cm^{3}$.
c. using the volume you calculated in (b), how many times must you fill the punch bowl during the party?
hint: convert your answer in (b) from $cm^{3}$ into ml first. remember $1 cm^{3}=1 ml$

Explanation:

Step1: Find the radius of the hemisphere

The diameter $d = 35$ cm, so the radius $r=\frac{d}{2}=\frac{35}{2}=17.5$ cm.

Step2: Calculate the volume of the hemisphere

The volume of a sphere is $V_{sphere}=\frac{4}{3}\pi r^{3}$. The volume of a hemisphere $V_{hemisphere}=\frac{1}{2}\times\frac{4}{3}\pi r^{3}=\frac{2}{3}\pi r^{3}$. Substitute $r = 17.5$ cm into the formula:
\[

$$\begin{align*} V_{hemisphere}&=\frac{2}{3}\times\pi\times(17.5)^{3}\\ &=\frac{2}{3}\times\pi\times5359.375\\ &\approx\frac{2\times3.14\times5359.375}{3}\\ &=\frac{33656.35}{3}\\ &\approx11218.78\text{ cm}^3 \end{align*}$$

\]

Step3: Calculate the 85% - capacity volume

Multiply the volume of the hemisphere by 0.85. $V = 0.85\times11218.78\approx9536\text{ cm}^3$.

Step4: Since part (c) lacks the total volume of punch needed for the party, we can't solve it. But for parts (a) and (b):

Answer:

a. $11218.78$ cm$^3$
b. $9536$ cm$^3$