91Ó°ÊÓ

The volume of the part of a ball lies between two parallel planes which intersect the ball.

The elementary volume in cylindrical coordinate system around a point$\mathit{P}\left(r,\mathrm{Ï•},z\right)$is given by$\mathit{d}\mathit{V}\mathbf{=}\mathit{r}\mathit{d}\mathit{r}\mathbf{}\mathbf{.}\mathit{d}\mathit{Ï•}\mathbf{}\mathbf{.}\mathit{d}\mathit{z}$.

(a)

Suppose be the radius of the ball, and planes $z=h_1,z=h_2$intersect the ball the shaded portion of the ball for which the volume integral/ triple integral is required is shown in the Figure below:

Consider point $P(\mathrm{r},\mathrm{Ï•},\mathrm{z})$within the required volume between in planes.

The volume element around point P is$dV=rdr.d\mathrm{ϕ}.dz$ where $0≤r≤\sqrt{R^2-z^2},0≤\mathrm{ϕ}≤2\mathrm{τ}\mathrm{τ}and{h}_1≤z≤h_2$.

Therefore, the required volume is $V=∫∫∫dV={∫}_0^{2\mathrm{τ}\mathrm{τ}}d\mathrm{ϕ}.{∫}_{h_1}^{h_2}\left({∫}_0^{\sqrt{R^2-z^2}}rdr\right)dz$.

That is the required triple integral in cylindrical coordinates for the volume of the part of a ball between two parallel planes which intersect the ball.

The required triple integral is $V=∫∫∫dV={∫}_0^{2\mathrm{τ}\mathrm{τ}}d\mathrm{ϕ}.{∫}_{h_1}^{h_2}\left({∫}_0^{\sqrt{R^2-z^2}}rdr\right)dz$.

(b)

The $I={∫}_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)\mathrm{where}∫f\left(x\right)dx\mathit{=}F\left(x\right)$.

The required volume is $V=∫∫∫dV={∫}_0^{2\mathrm{τ}\mathrm{τ}}d\mathrm{ϕ}.{∫}_{h_1}^{h_2}\left({∫}_0^{\sqrt{R^2-z^2}}rdr\right)dz$.

$$\begin{gathered} V={\left|\mathrm{ϕ}\right|}_0^{2\mathrm{τ}\mathrm{τ}}.{∫}_{h_1}^{h_2}\left({\left|\frac{r^2}{2}\right|}_0^{\sqrt{R^2-z^2}}\right)dz \\ V=2\mathrm{τ}\mathrm{τ}.\frac{1}{2}.{∫}_{h_1}^{h_2}\left(R^2-z^2\right)dz=\mathrm{τ}\mathrm{τ}{\left|R^{2}z-\frac{z^3}{3}\right|}_{h_1}^{h_2} \\ V=\mathrm{τ}\mathrm{τ}\left[R^2\left(h_2-h_1\right)-\frac{\left(h_2^3-h_1^3\right)}{3}\right] \end{gathered}$$

The solved integral in part (a) is $V=\mathrm{Ï„}\mathrm{Ï„}\left[R^2\left(h_2-h_1\right)-\frac{\left(h_2^3-h_1^3\right)}{3}\right]$.

(c)

The centroid of volume is the geometric centre of a body.

If the density of volume is uniform throughout the body, then the coordinates of the centroid of volume $G\left(\overline{x},\overline{y},\overline{z}\right)$is given as follows:

$$\begin{gathered} \overline{x}=\frac{∫xdV}{∫dV} \\ \overline{y}=\frac{∫ydV}{∫dV} \\ \overline{z}=\frac{∫zdV}{∫dV} \end{gathered}$$

As it is seen in Fig. in part (a), the volume of the ball between two planes is symmetrical about z-axis only.

Therefore, the volume of centroid lies on z-axis that is$G\left(0,0,\overline{z}\right).\overline{z}=\frac{∫zdV}{∫dV}=\frac{1}{V}∫zdV$

Calculate as follows:

role="math" localid="1659153662307" $$\begin{gathered} I={∫}_0^{2\mathrm{π}}df.{∫}_{h_1}^{h_2}\left({∫}_0^{\sqrt{R^2-z^2}}rdr\right)zdz \\ ={\left|f\right|}_0^{2\mathrm{π}}.{∫}_{h_1}^{h_2}\left({\left|\frac{r^2}{2}\right|}_0^{\sqrt{R^2-z^2}}\right)dz \\ 2\mathrm{π}.\frac{1}{2}{∫}_{h_1}^{h_2}\left(R^2-z^2\right)zdz=2\mathrm{π}.\frac{1}{2}{∫}_{h_1}^{h_2}\left(R^2-z^3\right)dz \\ =\mathrm{π}{\left|\frac{R^2{z}^2}{2}\frac{z^4}{4}\right|}_{h_1}^{h_2} \end{gathered}$$

Calculate further as follows:

$$\begin{gathered} I=\frac{\mathrm{Ï„}\mathrm{Ï„}}{2}\left[R^2\left(h_2^2-h_1^2\right)-\frac{1}{2}\left(h_2^4-h_1^4\right)\right] \\ \overline{z}=\frac{\mathrm{Ï„}\mathrm{Ï„}}{2V}\left[R^2\left(h_2^2-h_1^2\right)-\frac{1}{2}\left(h_2^4-h_1^4\right)\right] \\ \overline{z}=\frac{\left[R^2\left(h_2^2-h_1^2\right)-\frac{1}{2}\left(h_2^4-h_1^4\right)\right]}{2\left[R^2\left(h_2^2-h_1^2\right)-\frac{1}{3}\left(h_2^3-h_1^3\right)\right]} \end{gathered}$$

Thus, the centroid of the volume described in part (a) is.

$\left(0,0,\overline{z}=\frac{R^2\left(h_2^2-h_1^2\right)-\left(h_2^4-h_1^4\right)/2}{2R^2\left(h_2-h_1\right)-2\left(h_2^3-h_1^3\right)/3}\right)$

The centroid of the volume described in part (a) is $\left(0,0,\overline{z}=\frac{R^2\left(h_2^2-h_1^2\right)-\left(h_2^4-h_1^4\right)/2}{2R^2\left(h_2-h_1\right)-2\left(h_2^3-h_1^3\right)/3}\right)$.