Make D as a term of a.
1/3πD/2²D = 4/3πa³
Volume of a cone where h is D and radius is D/2
Volume of a sphere with a as the radius

Question

Make D as a term of a. 1/3πD/2²D = 4/3πa³ Volume of a cone where h is D and radius is D/2 Volume of a sphere with a as the radius

SnapAndSolve AI · Accepted Answer

Sure, let's solve for D in terms of a using the given equations.
First, let's start with the volume of the cone where the height (h) is D and the radius (r) is D/2. The formula for the volume of a cone is:
\[
V_{cone} = \frac{1}{3} \pi r^2 h
\]
Substituting the values of r and h:
\[
V_{cone} = \frac{1}{3} \pi \left(\frac{D}{2}\right)^2 D
\]
\[
V_{cone} = \frac{1}{3} \pi \left(\frac{D^2}{4}\right) D
\]
\[
V_{cone} = \frac{1}{3} \pi \frac{D^3}{4}
\]
\[
V_{cone} = \frac{1}{12} \pi D^3
\]
Next, let's calculate the volume of a sphere with radius a, which is given by:
\[
V_{sphere} = \frac{4}{3} \pi a^3
\]
Setting the volumes of the cone and sphere equal to each other as given by the equation:
\[
\frac{1}{12} \pi D^3 = \frac{4}{3} \pi a^3
\]
Now, we can simplify this equation by canceling π from both sides:
\[
\frac{1}{12} D^3 = \frac{4}{3} a^3
\]
To eliminate the fraction, multiply both sides by 12:
\[
D^3 = 12 \cdot \frac{4}{3} a^3
\]
\[
D^3 = 12 \cdot \frac{4}{3} a^3
\]
\[
D^3 = 16a^3
\]
Next, we will solve for D by taking the cube root of both sides:
\[
D = \sqrt[3]{16a^3}
\]
This can be further simplified:
\[
D = a \sqrt[3]{16}
\]
Since \( \sqrt[3]{16} \) can be expressed as \( 2 \sqrt[3]{2} \):
\[
D = 2 a \sqrt[3]{2}
\]
Now, we have the expression for D in terms of a:

D
=
2
a

2

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Make D as a term of a. 1/3πD/2²D = 4/3πa³ Volume of a cone where h is D and radius is D/2 Volume of a...

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