Question

(a) Find the centroid of the solid paraboloid inside

$\mathbf{z}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{0}\mathbf{<}\mathbf{z}\mathbf{<}\mathbf{c}$

(b) Repeat part (a) if the density is$\mathbf{\rho }\mathbf{=}\mathbf{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$

Short answer

The centroid of the solid paraboloid are as follows:

$$\begin{gathered} \left(\overline{x},\overline{y},\overline{z}\right)=\left(0,0,\frac{2}{3}c\right) \\ \left(\overline{x},\overline{y},\overline{z}\right)=\left(0,0,\frac{5}{7}c\right) \end{gathered}$$

Step-by-step solution

Given Condition

The solid paraboloid is$z=x^2+y^2,0<z<c$

The density is $\rho =r=\sqrt{x^2+y^2}$

Concept of centroid

The centroid or geometric center of a plane figure is the arithmetic mean position of all the points in the figure.

Draw the diagram

(a) Find the centroid of the solid paraboloid inside$\mathbf{z}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{0}\mathbf{<}\mathbf{z}\mathbf{<}\mathbf{c}$

Calculate the volume.

The volume is as given below,

$$\begin{gathered} V={\int }_0^{2\mathrm{\pi }}d\theta {\int }_0^{c}dz{\int }_0^{\sqrt{z}}rdr \\ =2\mathrm{\pi }{\int }_0^{\mathrm{c}}\mathrm{dz}\frac{1}{2}\mathrm{z} \\ =\frac{\mathrm{\pi }}{2}{\mathrm{c}}^2 \\ \overline{\mathrm{x}}=\overline{\mathrm{y}}=0 \end{gathered}$$

Thus, to find$\overline{z}$ .

 Step 5: Calculate the coordinate.

The value of $\overline{z}$ is as follows,

$$\begin{gathered} V={\int }_0^{2\mathrm{\pi }}d\theta {\int }_0^{c}dz{\int }_0^{\sqrt{z}}rdr \\ =2\mathrm{\pi }{\int }_0^{\mathrm{c}}\mathrm{dz}\frac{1}{2}\mathrm{z} \\ =\frac{\mathrm{\pi }}{2}{\mathrm{c}}^3 \\ \overline{z}=\frac{2}{3}c \end{gathered}$$

Hence, the centroid is $\left(\overline{x},\overline{y},\overline{z}\right)=\left(0,0\frac{2}{3}c\right)$

(b) Repeat part (a) if the density is $\mathbf{\rho }\mathbf{=}\mathbf{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$

Calculate the center of mass.

To find center of mass. Assume that if the density function changes, the centroid does not change.

Therefore, the mass is given below.

$$\begin{gathered} M={\int }_0^{2\mathrm{\pi }}d\theta {\int }_0^{c}dz{\int }_0^{\sqrt{z}}rdr\rho \\ ={\int }_0^{2\mathrm{\pi }}d\theta {\int }_0^{c}dz{\int }_0^{\sqrt{z}}{r}^{2}dr \\ =\frac{2\mathrm{\pi }}{3}{\int }_0^{c}dz{z}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {\overline{)=\frac{2\mathrm{\pi }}{3}\frac{2}{5}{z}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}_0^c \\ =\frac{4\mathrm{\pi }}{15}{c}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Calculate the coordinate.

On getting, $\overline{x}=\overline{y}=0$, the only necessity is to find $\overline{z}$.

Hence, the centroid is $\left(\overline{x},\overline{y},\overline{z}\right)=\left(0,0,\frac{5}{7}c\right)$