Q. 62 - Calculus

Question

Use a definite integral to prove that the volume formula $V=\frac{4}{3}{\mathrm{\pi r}}^3$ holds for a sphere of radius 3. 

Step-by-Step Solution

Answer

The volume of the sphere is $36\mathrm{\pi }$ by suing formula and integration also.

Step 1. Given information

The formula for the volume of sphere is $V=\frac{4}{3}{\mathrm{\pi r}}^3$, where $r$ is tthe radius

The radius is 3.

Step 2. Find volume of sphere using formula:

$$\begin{gathered} V=\frac{4}{3}\mathrm{\pi }{\left(3\right)}^3 \\ =36\mathrm{\pi } \end{gathered}$$

Step 3. Volume of sphere using integration

Draw a sphere of radius 3:

A circular slice of thickness $dx$ has been cut at a distance of $x$ from the center of the sphere.

The radius of the circular slice is:

$AB=\sqrt{3^2-x^2}$

So the area of the slice is:$$\begin{gathered} A=\mathrm{\pi }{\left(\mathrm{AB}\right)}^2 \\ =\mathrm{\pi }(3^2-{\mathrm{x}}^2) \\ =\mathrm{\pi }(9-{\mathrm{x}}^2) \end{gathered}$$

The volume of the slice is $$\begin{gathered} dV=Adx \\ =\mathrm{\pi }\left(9-{\mathrm{x}}^2\right)\mathrm{dx} \end{gathered}$$

Hence the volume of the sphere is:

$$\begin{gathered} V={\int }_{-3}^{3}dV \\ ={\int }_{-3}^3\mathrm{\pi }\left(9-{\mathrm{x}}^2\right)\mathrm{dx} \\ =\mathrm{\pi }{\int }_{-3}^3\left(9-{\mathrm{x}}^2\right)\mathrm{dx} \\ =2\mathrm{\pi }{\int }_0^3\left(9-{\mathrm{x}}^2\right)\mathrm{dx} \\ =2\mathrm{\pi }{\left[9\mathrm{x}-\frac{{\mathrm{x}}^3}{3}\right]}_0^3 \\ =2\mathrm{\pi }\left(9\times 3-\frac{3^3}{3}-0\right) \\ =2\mathrm{\pi }\left(27-9\right) \\ =36\mathrm{\pi } \end{gathered}$$