Question

As needed, use a computer to plot graphs and to check values of integrals. (a) Find the centroid of the solid paraboloid inside $z=x^2+y^2,0<z<c$. (b) Repeat part (a) if the density is $\rho =r=\sqrt{x^2+y^2}.$

Short answer

(a) $(0,0,\frac{c}{3})$. (b) $(0,0,\frac{5c}{14})$.

Step-by-step solution

Set up the integral for the volume

The volume of the solid paraboloid can be found by setting up the triple integral in cylindrical coordinates. Use the transformation: $x=r\cos \theta$, $y=r\sin \theta$, $z=z$. The limits are: $0\le r\le \sqrt{c}$, $0\le \theta \le 2\pi$, and $0\le z\le r^2$. The volume element in cylindrical coordinates is $rdrd\theta dz$.

Calculate the volume

The volume $V$ is given by the integral: $$V={\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}rdzdrd\theta$$. Evaluate the innermost integral first with respect to $z$.

Evaluate the inner integral

$${\int }_0^{r^2}rdz=r[z{]}_0^{r^2}=r(r^2)=r^3$$

Evaluate the next integral

Now integrate with respect to $r$: $${\int }_0^{\sqrt{c}}{r}^{3}dr$$. This gives: $${\int }_0^{\sqrt{c}}{r}^{3}dr={\left[\frac{r^4}{4}\right]}_0^{\sqrt{c}}=\frac{(\sqrt{c}{)}^4}{4}=\frac{c^2}{4}$$

Evaluate the outer integral

Finally, integrate with respect to $\theta$: $${\int }_0^{2\pi }\frac{c^2}{4}d\theta =\frac{c^2}{4}[\theta {]}_0^{2\pi }=\frac{c^2}{4}\times 2\pi =\frac{\pi c^2}{2}$$

Set up integrals for the coordinates of the centroid

The centroid $\overline{x},\overline{y},\overline{z}$ is given by: $\overline{x}=\frac{1}{V}\int \int \int xdV$, $\overline{y}=\frac{1}{V}\int \int \int ydV$, and $\overline{z}=\frac{1}{V}\int \int \int zdV$. By symmetry, $\overline{x}=\overline{y}=0$. Therefore, we only need to find $\overline{z}$.

Calculate $\overline{z}$

The integral for $\overline{z}$ is: $$\overline{z}=\frac{1}{V}{\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}zrdzdrd\theta$$. Evaluate the innermost integral first: $${\int }_0^{r^2}zrdz=r{\left[\frac{z^2}{2}\right]}_0^{r^2}=r\frac{(r^2{)}^2}{2}=\frac{r^5}{2}$$

Evaluate the next integral

$${\int }_0^{\sqrt{c}}\frac{r^5}{2}dr=\frac{1}{2}{\int }_0^{\sqrt{c}}{r}^{5}dr=\frac{1}{2}{\left[\frac{r^6}{6}\right]}_0^{\sqrt{c}}=\frac{1}{2}\times \frac{(\sqrt{c}{)}^6}{6}=\frac{c^3}{12}$$

Evaluate the outer integral

$${\int }_0^{2\pi }\frac{c^3}{12}d\theta =\frac{c^3}{12}{\left[\theta \right]}_0^{2\pi }=\frac{c^3}{12}\times 2\pi =\frac{\pi c^3}{6}$$

Compute $\overline{z}$

$$\overline{z}=\frac{1}{V}{\int }_0^{2\pi }{\int }_0^{\sqrt{c}}\frac{r^5}{2}drd\theta =\frac{1}{\frac{\pi c^2}{2}}\times \frac{\pi c^3}{6}=\frac{2}{c^2}\times \frac{\pi c^3}{6}=\frac{c}{3}$$

Answer for Part (a)

The centroid of the solid paraboloid is: $(0,0,\frac{c}{3})$.

Set up the integral for the mass

For part (b), the density $\rho =r$. The mass $M$ is given by: $$M={\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}r\cdot rdzdrd\theta ={\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}{r}^{2}dzdrd\theta$$

Evaluate the inner integral for mass

$${\int }_0^{r^2}{r}^{2}dz=r^2[z{]}_0^{r^2}=r^2\times r^2=r^4$$

Evaluate the next integral for mass

$${\int }_0^{\sqrt{c}}{r}^{4}dr={\left[\frac{r^5}{5}\right]}_0^{\sqrt{c}}=\frac{(\sqrt{c}{)}^5}{5}=\frac{c^{5/2}}{5}$$

Evaluate the outer integral for mass

$${\int }_0^{2\pi }\frac{c^{5/2}}{5}d\theta =\frac{c^{5/2}}{5}{\left[\theta \right]}_0^{2\pi }=\frac{c^{5/2}}{5}\times 2\pi =\frac{2\pi c^{5/2}}{5}$$

Set up the integrals for the coordinates of the centroid with density

For $\overline{z}$ with density $\rho =r$, calculate: $$\overline{z}=\frac{1}{M}{\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}z{r}^{2}dzdrd\theta$$.

Evaluate the inner integral for $\overline{z}$ with density

$${\int }_0^{r^2}z{r}^{2}dz=r^2{\left[\frac{z^2}{2}\right]}_0^{r^2}=r^2\frac{(r^2{)}^2}{2}=\frac{r^6}{2}$$

Evaluate the next integral for $\overline{z}$ with density

$${\int }_0^{\sqrt{c}}\frac{r^6}{2}dr=\frac{1}{2}{\int }_0^{\sqrt{c}}{r}^{6}dr=\frac{1}{2}{\left[\frac{r^7}{7}\right]}_0^{\sqrt{c}}=\frac{1}{2}\times \frac{(\sqrt{c}{)}^7}{7}=\frac{c^{7/2}}{14}$$

Evaluate the outer integral for $\overline{z}$ with density

$${\int }_0^{2\pi }\frac{c^{7/2}}{14}d\theta =\frac{c^{7/2}}{14}{\left[\theta \right]}_0^{2\pi }=\frac{c^{7/2}}{14}\times 2\pi =\frac{\pi c^{7/2}}{7}$$

Compute $\overline{z}$ with density

$$\overline{z}=\frac{1}{M}{\int }_0^{2\pi }{\int }_0^{\sqrt{c}}\frac{r^6}{2}drd\theta =\frac{1}{\frac{2\pi c^{5/2}}{5}}\times \frac{\pi c^{7/2}}{7}=\frac{5}{2\pi c^{5/2}}\times \frac{\pi c^{7/2}}{7}=\frac{5c}{14}$$

Answer for Part (b)

The centroid of the solid paraboloid with the given density is: $(0,0,\frac{5c}{14})$.

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional extension of polar coordinates. They are particularly useful for problems with rotational symmetry around an axis. In cylindrical coordinates, a point in space is represented by three values: $(r,\theta ,z)$. Here’s what each term means:

\* \textbf{r} is the distance from the z-axis (similar to radius in polar coordinates).
\* \textbf{θ} (theta) is the angle measured from the positive x-axis in the xy-plane.
\* \textbf{z} is the height above the xy-plane.

The transformation from Cartesian coordinates $x,y,z$ to cylindrical coordinates is given by: $x=r\cos \theta ,y=r\sin \theta ,z=z$.
This system simplifies the integration process for solids of revolution, such as cylinders, cones, and paraboloids, allowing us to exploit their inherent symmetries.

Triple Integral

A triple integral extends the concept of a double integral to three dimensions. It integrates a function over a volume. When we want to calculate the volume of a solid or the mass with a given density, we use triple integrals.

For a function $f(x,y,z)$, the triple integral over a volume $V$ is written as:
$\int \int {\int }_{V}f(x,y,z)dV$

In cylindrical coordinates, the volume element $dV$ is represented as $rdrd\theta dz$, making it easier to handle integrals involving rotational symmetries.
The limits of the triple integral are determined based on the geometry of the solid being considered. For instance, the bounds for $r$ range from 0 to the radial boundary, $\theta$ ranges from 0 to $2\pi$, and $z$ ranges from the bottom to the top surface of the volume.

Volume Calculation

Volume calculation using integration helps us determine the space occupied by a solid. For the solid paraboloid defined by $z=x^2+y^2$, integrated from $z=0$ to $z=c$, we perform the triple integral:
$$V={\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}rdzdrd\theta$$
Steps to solve the integral:

\1. Calculate the innermost integral with respect to $z$.
${\int }_0^{r^2}rdz=r[z{]}_0^{r^2}=r(r^2)=r^3$.
\2. Integrate the result with respect to $r$.
${\int }_0^{\sqrt{c}}{r}^{3}dr=\text{[}\frac{r^4}{4}{\text{]}}_0^{\sqrt{c}}=\frac{c^2}{4}$.
\3. Finally, integrate with respect to $\theta$.
${\int }_0^{2\pi }\frac{c^2}{4}d\theta =\frac{c^2}{4}[\theta {]}_0^{2\pi }=\frac{c^2}{4}\times 2\pi =\frac{\pi c^2}{2}$. The volume $V$ of the solid paraboloid is $\frac{\pi c^2}{2}$.

Mass and Density

To find the mass of a solid with a varying density, integrate the density function over the volume.

For the given solid paraboloid with density $\rho =r$:
$M={\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}r\cdot rdzdrd\theta$.
Substitute and simplify to get:
$$M={\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}{r}^{2}dzdrd\theta$$
The inner integral becomes:
$${\int }_0^{r^2}{r}^{2}dz=r^2[z{]}_0^{r^2}=r^2\times r^2=r^4$$
Next integral:
$${\int }_0^{\sqrt{c}}{r}^{4}dr=\text{[}\frac{r^5}{5}$$_0^{\sqrt{c}} = \frac{(\sqrt{c})^5}{5} = \frac{c^{5/2}}{5} \]
Finally:
$${\int }_0^{2\pi }\frac{c^{5/2}}{5}d\theta =\frac{c^{5/2}}{5}[\theta {]}_0^{2\pi }=\frac{c^{5/2}}{5}\times 2\pi =\frac{2\pi c^{5/2}}{5}$$
Thus, the mass $M$ is $\frac{2\pi c^{5/2}}{5}$.
Next, to find $\overline{z}$ for the centroid, integrate $z\rho$ over the volume and normalize by mass. The integral:
$$\overline{z}=\frac{1}{M}{\int }_0^{2\pi }{\int }_0^{\sqrt{c}}{\int }_0^{r^2}z{r}^{2}dzdrd\theta$$
Perform the inner integral:
$${\int }_0^{r^2}z{r}^{2}dz=r^2{\left[\frac{z^2}{2}\right]}_0^{r^2}=r^2\frac{r^4}{2}=\frac{r^6}{2}$$
Then:
$${\int }_0^{\sqrt{c}}\frac{r^6}{2}dr=\frac{1}{2}{\int }_0^{\sqrt{c}}{r}^{6}dr=\text{[}\frac{r^7}{7}$$_0^{\sqrt{c}} = \frac{c^{7/2}}{14} \]
Finally:
$${\int }_0^{2\pi }\frac{c^{7/2}}{14}d\theta =\frac{c^{7/2}}{14}[\theta {]}_0^{2\pi }=\frac{c^{7/2}}{14}\times 2\pi =\frac{\pi c^{7/2}}{7}$$
Thus:
$\overline{z}=\frac{5c}{14}$. Hence, the centroid is $(0,0,\frac{5c}{14})$.