Center of mass and moments

 The expressions of mass, center of mass, moments of inertia expressed in double integrals can be modified by replacing the double integrals with triple integrals.

Example

Solution

The region Q is a tetrahedron meeting the axes at the point (6,0,0), (0,3,0) and (0,0,2) (see figure below). To find the limits of integration, let z = 0 in the slanted plane z = 1/3(6-x-2y). Then for x and y find the projection of Q on the the xy plane which is bounded by the axes and the line x_+ 2y = 6. The mass is calculated as follow:

 

Finding the moments of inertia of a solid in two dimensions.

 Goal: Find the moment of inertia of a solid in two dimensions

Let's go back to the lamina considered in earlier post where we calculated its mass. In order to do that we considered the region R occupied by the lamina. We divided the lamina into tiny subrectangles. Our goal now is to find the different moments of inertia of the solid.

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Let's find the moment of inertia about the x-axis.

 The moment of inertia of a solid about an axis is equal to its mass by the square of its distance from this axis.

The moment of inertia of the lamina about the x-axis is equal to the sum of the moments of the tiny subrectangles of the lamina.

The moment of inertia of a subrectangle about the x-axis is equal to the product of its mass by the square of its distance from the x-axis. This moment is calculated as: 

Let's add all the moments of the subrectangles. This comes to take the Rieman sum of the product and to determine its limit.                                                                                                                                         

           

The moment of inertia of a subrectangle about the y-axis is equal to the product of its mass by the square of its distance from the t-axis.  Hence, the moment is given by:                                                                      

                                                                                                                     

Adding all the moments together allows to find the moment of the lamina about the y-axis. This leads to use the Rieman sum of the product and take its limit.

Example

Solution

Practice

Center of mass in two dimensions

 The center of mass of an object is called the center of gravity if the object is located in a gravitational field. If the object has a uniform density, the center of mass is the geometric center of the object, which is called the centroid. The following figure shows the point P as the center of mass of a lamina (flat plate).

 

Restating the center of mass in terms of integrals, we have:

If the object has uniform density, the density function ⍴(x,y) is constant and the formulas become:

Example

Solution

Practice

Application of double integrals: finding moments about x and y axes

 Objective: Finding moments about x and y axis

Let's find the moments with respect to x and y of a region R with density function ⍴(x,y). We divide the region in small rectangles for which the density is constant. We add the moments of each of these rectangles and take the limit of the sum as as the rectangles approach zero.

If we name Rᵢⱼ the small rectangle, the moment with respect to the x-axis is:

Similarly, the moment with respect to the y-axis is defined by:

Example

We use the same example in a previous blog post

Solution

Practice

Applications of double Integrals: Mass of an object in a two-dimensional space.

 Let's consider a lamina (a thin plate) that occupies the region R of a two-dimensional space. Let (x,y) be a point in the region R surrounded by a small rectangle. The density of the lamina at that point is a function ⍴(x,y). This function is determined by: 

where 𝚫m and 𝚫A represent the mass and the area of the small rectangle.

Let's divide the region R into tiny rectangles of area ΔA. The mass of a tiny rectangle is given by:

Let's add the tiny rectangles together and take the limit of the sum when delta x and delta y approach zero. Since we have two variables, we know by experience that the limit is a double sum of Rieman. This limit represents a double integral that allows us to find the mass off the lamina.

Example

Solution

Let's sketch the region R:

Practice 

Find the volume inside of an ellipsoid and outside of a sphere

 We are going to use an example to do that:

Example:

Solution

First let's find the volume using a = 75ft b = 80 ft c = 90ft using the result from the example treated previously about the volume of the ellipsoid. Hence the volume of the ellipsoid is: 

From the result from the example about the volume of a sphere, we get:

Volume of an ellipsoid using spherical coordinates

 Let's use an example to calculate the volume of an ellipsoid using spherical coordinates.

Example

Solution

Let's use the change of variables that corresponds to an ellipsoid. We still change the variables from rectangular coordinates to spherical coordinates. 

The volume V of the ellipsoid is given by:

Let's change dV = dxdydz in spherical coordinates:

Let's apply the change-of-variables formula

$$dx\,dy\,dz=\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right|\;d\rho\,d\theta\,d\varphi.$$

$$quotient\; of\; the\; partial\; derivatives.$$

So, the task is to compute the Jacobian:

$$J(\rho,\theta,\varphi)=\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}.$$

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Compute the partial derivatives

Differentiate $x,y,\mathrm{z\; with\; respect\; to }$$\rho,\theta,\varphi$.

With respect to $\rho$:

$$\frac{\partial(x,y,z)}{\partial\rho} =\Big(a\cos\varphi\sin\theta,\; b\sin\varphi\sin\theta,\; c\cos\theta\Big).$$

With respect to $\theta$:

$$\frac{\partial(x,y,z)}{\partial\theta} =\Big(a\rho\cos\varphi\cos\theta,\; b\rho\sin\varphi\cos\theta,\; -c\rho\sin\theta\Big).$$

With respect to $\varphi$:

$$\frac{\partial(x,y,z)}{\partial\varphi} =\Big(-a\rho\sin\varphi\sin\theta,\; b\rho\cos\varphi\sin\theta,\; 0\Big).$$

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 Form the Jacobian determinant

Put those three vectors as columns (or rows—just be consistent). Using columns:

$$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| = \left| \begin{matrix} a\cos\varphi\sin\theta & a\rho\cos\varphi\cos\theta & -a\rho\sin\varphi\sin\theta\\ b\sin\varphi\sin\theta & b\rho\sin\varphi\cos\theta & \ \ b\rho\cos\varphi\sin\theta\\ c\cos\theta & -c\rho\sin\theta & 0 \end{matrix} \right|.$$

 Factor constants (the key simplification)

• Factor $a$ from row 1, $b$ from row 2, $c$ from row 3 $<br>$

• Factor $\rho$ from column 2 and $\rho$ from column 3 $<br>$

So

$$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| = abc\,\rho^2 \left| \begin{matrix} \cos\varphi\sin\theta & \cos\varphi\cos\theta & -\sin\varphi\sin\theta\\ \sin\varphi\sin\theta & \sin\varphi\cos\theta & \ \cos\varphi\sin\theta\\ \cos\theta & -\sin\theta & 0 \end{matrix} \right|.$$ 

Evaluating the determinant we obtain:

$$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| =abc\,\rho^2\sin\theta.$$ 

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Final conversion of dV

$$\boxed{dx\,dy\,dz=abc\,\rho^2\sin\theta\;d\rho\,d\theta\,d\varphi.}$$

The volume of the ellipsoid is calculated as follows:

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