# Volume of Shapes

I'm doing some maths revision at home from a GSCE book and I'm a bit stuck. I'm not sure how to start off. (I know the formulae for a volume of cone and a formulae for a volume of a sphere but I don't know how that will help me).

QUESTION:

If a cone with perpendicular height $6h$ and radius $2h$ has the same volume as a sphere of radius $r$, show that $r = \sqrt[3]{6h}$.

• Type \$\sqrt{x+y}\$ to get $\sqrt{x+y}$ or \$\root {m+n} \of {x+y}\$to get $\root {m+n} \of {x+y}$. – Git Gud Jun 27 '13 at 17:38
• Appending to @GitGud: another approach is $\sqrt[m+n]{x+y}$ for $\sqrt[m+n]{x+y}$ – apnorton Jun 27 '13 at 17:42
• @anorton Yours is easier to type. Thanks. – Git Gud Jun 27 '13 at 17:43
• Note that your equation is dimensionally inconsistent (in a physical sense). It should be $r=\sqrt[3]{6h^3}$. – pppqqq Jun 27 '13 at 18:05

The volume of a cone is $\frac{1}{3}\pi (\rm{radius})^2(\rm{height})$, which in this case is $\frac{1}{3}\pi (2h)^2(6h) = 8\pi h^3$.

The volume of a sphere is $\frac{4}{3}\pi (\rm{radius})^3$, which in this case is $\frac{4}{3}\pi r^3$.

You're told that these volumes are equal: $8\pi h^3 = \frac{4}{3} \pi r^3$. Can you see what to do from here?

From the formulas for the area of a circle and cone: $$V_{\text{circ}} = \frac{4}{3}\pi r^3$$ $$V_{\text{cone}} = \frac{1}{3}\pi \underbrace{(2h)^2}_{\text{cone radius}}\cdot\underbrace{(6h)}_{\text{cone height}}$$

We are told that the volumes are the same. Thus, we can set: $$V_{\text{circ}} =V_{\text{cone}}$$ Thus: $$\frac{4}{3}\pi r^3 = \frac{1}{3}\pi(2h)^2\cdot(6h)$$

Simplifying the $(2h)^2$: $$\frac{4}{3}\pi r^3 = \frac{1}{3}\pi(4r^2)(6h)$$ Now, multiply by $\frac{3}{4}$: $$\pi r^3 = \pi(r^2)(6h)$$

Volume of cone $$V_1=\dfrac 13 \pi R^2 H$$ R=2h,H=6h $$V_1=\dfrac 13 \pi {(2h)}^2 6h$$
Volume of sphere $$V_2=\dfrac 43 \pi r^3$$ but $$V_1=V_2$$ $$\dfrac 13 \pi {(2h)}^2 6h=\dfrac 43 \pi r^3$$ $$8\pi h^3=\dfrac 43 \pi r^3$$ $$r^3=\dfrac{24h^3\pi}{4\pi}$$ $$r^3=6h^3$$ $$r=\sqrt[3]{6}h$$