Q. 61 - Calculus

Question

Use a definite integral to prove that the volume formula $V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$ holds for a cone of radius 3 and height 5. 

Step-by-Step Solution

Answer

The volume of the cone is $15\mathrm{\pi }$ cubic unit and it is proved by definite integral.

Step 1. Given information

Height of the cone is $5$ and radius is $3$

Formula of volume of the cone is $V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Step 2. Find volume of the cone by using given formula:

$$\begin{gathered} V=\frac{1}{3}\mathrm{\pi }{\left(3\right)}^2\left(5\right) \\ =15\mathrm{\pi } \end{gathered}$$

Step 3. To prove the volume using integration.

Draw a 3d cone of the given dimension;

Here a circular slice has been cut at a  distance of $x$ from the base of the cone whose thickness is very small: $dx$

So the radius of the slice is calculated as:

Since, $∆ABC~∆ADE$

Hence, $$\begin{gathered} \frac{BC}{3}=\frac{5-x}{5} \\ BC=\frac{3}{5}(5-x) \\ =3-\frac{3}{5}x \end{gathered}$$

Area of the slice is:$$\begin{gathered} A=\mathrm{\pi }{\left(\mathrm{BC}\right)}^2=\mathrm{\pi }{\left(3-\frac{3}{5}\mathrm{x}\right)}^2 \\ =\mathrm{\pi }\left(9+\frac{9}{25}{\mathrm{x}}^2-\frac{18}{5}\mathrm{x}\right) \end{gathered}$$

Volume of the slice:

$dv=\mathrm{\pi }\left(9+\frac{9}{25{\mathrm{x}}^2}-\frac{18}{5}\mathrm{x}\right)\mathrm{dx}$

So volume of the cone is:

$$\begin{gathered} V={\int }_0^{5}dv \\ ={\int }_0^5\mathrm{\pi }\left(9+\frac{9}{25}{\mathrm{x}}^2-\frac{18}{5}\mathrm{x}\right)\mathrm{dx} \\ =\mathrm{\pi }{{\left[9\mathrm{x}+\frac{9{\mathrm{x}}^3}{75}-\frac{18{\mathrm{x}}^2}{10}\right]}^5}_0 \\ =\mathrm{\pi }\left[\left(9\times 5+\frac{9\times 5^3}{75}-\frac{18\times 5^2}{10}\right)-0\right] \\ =\mathrm{\pi }\left(45+15-45\right) \\ =15\mathrm{\pi } \end{gathered}$$