A sphere and a cylinder have equal volumes. The sphere has a radius 3r. The cylinder has radius 2r and height h. Find h in terms of r.
I'm only 15, someone walk me through this as simple as possible :)
$\endgroup$ 12 Answers
$\begingroup$Volume of sphere $= \frac 43 \pi R^3 = \frac 43 \pi (3r)^3$
Volume of cylinder $= \pi R^2 h = \pi(2r)^2h$
These two volumes should be equal so equate them:
$$\frac 43 \pi (3r)^3 = \pi(2r)^2h$$
Now let's solve for $h$ in terms of $r$.
First notice that $(3r)^3 = 3^3r^3 = 27r^3$. Likewise $(2r)^2 = 2^2r^2 = 4r^2$. Therefore
$$\frac 43 \pi (3r)^3 = \pi(2r)^2h \\ \implies \frac 43 \pi \color{red}{(27r^3)} = \pi\color{red}{(4r^2)}h \\ \implies \frac 43 \color{red}{(27r^3)}\pi = \color{red}{(4r^2)}\pi h \\ \implies \color{blue}{\frac{4\cdot 27}{3}}r^3\pi = 4r^2\pi h$$
Now before we go on, notice that there is a $4r^2\pi$ on both sides. So let's divide both sides by $4r^2\pi$:
$$\require{cancel}\frac{4\cdot 27}{3}r^3\pi = 4r^2\pi h \\ \implies \frac {27}{3}\color{blue}{(4r^2\pi)}r = \color{blue}{(4r^2\pi)} h \\ \implies \frac{\frac {27}{3}\cancel{\color{blue}{(4r^2\pi)}}r}{\cancel{\color{blue}{(4r^2\pi)}}} = \frac{\cancel{\color{blue}{(4r^2\pi)}} h}{\cancel{\color{blue}{(4r^2\pi)}}} \\ \implies \frac {27}{3}r=h \\ \implies 9r=h$$
So your answer is $\require{enclose}\enclose{box} {h=9r}$. This is an expression for $h$ in terms of $r$.
$\endgroup$ 11 $\begingroup$So $\frac{4}{3}\pi (3r)^3 = h\pi(2r)^2 \implies \pi36r^3 = h\pi4r^2 \implies 9r = h$.
$\endgroup$
