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What is the volume of a hemisphere with a diameter of
8.6ft, rounded to the nearest tenth of a cubic foot?
Answer:
ft^(3)

What is the volume of a hemisphere with a diameter of $8.6 \mathrm{ft}$ , rounded to the nearest tenth of a cubic foot? $\newline$ Answer: $\mathrm{ft}^{3}$

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Convert between customary and metric systems

Full solution

Q. What is the volume of a hemisphere with a diameter of

8.6 \mathrm{ft}

, rounded to the nearest tenth of a cubic foot?

\newline

Answer:

\mathrm{ft}^{3}

Calculate Sphere Volume: To find the volume of a hemisphere, we first need to find the volume of a full sphere and then divide it by $2$ . The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$ , where $r$ is the radius of the sphere.
Find Hemisphere Radius: The diameter of the hemisphere is given as $8.6$ feet. The radius $r$ is half of the diameter, so we need to divide $8.6$ by $2$ to find the radius. $\newline$ $r = \frac{8.6 \text{ ft}}{2} = 4.3 \text{ ft}$
Substitute Radius in Formula: Now we can substitute the radius into the volume formula for a sphere and then divide by $2$ to find the volume of the hemisphere. $\newline$ $V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi r^3$ $\newline$ $V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi(4.3 \text{ ft})^3$
Calculate Volume: Let's calculate the volume using the values we have: $\newline$ $V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi(4.3 \, \text{ft})^3$ $\newline$ $V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi(79.507 \, \text{ft}^3)$ (since $4.3^3 = 79.507$ )
Perform Multiplication: Now we perform the multiplication: $\newline$ $V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \times \pi \times 79.507 \, \text{ft}^3$ $\newline$ $V_{\text{hemisphere}} = \frac{2}{3} \times \pi \times 79.507 \, \text{ft}^3$
Use Approximate Value for $\pi$ : We can now use the approximate value for $\pi$ $3.14159$ to calculate the volume: $V_{\text{hemisphere}} = \left(\frac{2}{3}\right) \times 3.14159 \times 79.507 \, \text{ft}^3$ $V_{\text{hemisphere}} \approx 167.551 \, \text{ft}^3$
Round to Nearest Tenth: Finally, we round the volume to the nearest tenth of a cubic foot as requested: $V_{\text{hemisphere}} \approx 167.6 \, \text{ft}^3$

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