How to Find Mass and Center of Mass for a Triangle with Variable Vertices?

[harpazo]

Find the mass and center of mass given the following region R.

R: triangle with vertices (0,0), (b/2, h), (b,0)

Let p = rho

p = k

The letters b and h as given in the points make it hard to find the equation of the lines needed for the inner upper and lower limits. I have worked this out several times but get stuck somewhere alone the way. I need help with the set up. I can take it from there.

 

[MarkFL]

Suppose we have two points in the plane:

$$P_1\left(x_1,y_1\right)$$

$$P_2\left(x_2,y_2\right)$$

Then, by the definition of slope, and using the point-slope formula, the equation of the line through the two points is:

$$y=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)+y_1$$

Using this formula, can you find the equations of the lines along which the boundaries of the triangle lie?

 

[harpazo]

Suppose we have two points in the plane:

$$P_1\left(x_1,y_1\right)$$

$$P_2\left(x_2,y_2\right)$$

Then, by the definition of slope, and using the point-slope formula, the equation of the line through the two points is:

$$y=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)+y_1$$

Using this formula, can you find the equations of the lines along which the boundaries of the triangle lie?

Can you find one of the equations for me showing every step? Also, once I have the two equations, do I need to solve for x if integrating over dxdy? Should I integrate over dydx instead? Which is easier set up and why?

 

[harpazo]

Suppose we have two points in the plane:

$$P_1\left(x_1,y_1\right)$$

$$P_2\left(x_2,y_2\right)$$

Then, by the definition of slope, and using the point-slope formula, the equation of the line through the two points is:

$$y=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)+y_1$$

Using this formula, can you find the equations of the lines along which the boundaries of the triangle lie?

Using the formula you provided, I found y = (2h/b)x to be one of the equations and the other equation I found to be y = (-2h/b)(x - b/2) + h.

Is any of this right? I know my college algebra is a bit rusty.

 

[MarkFL]

Using the formula you provided, I found y = (2h/b)x to be one of the equations and the other equation I found to be y = (-2h/b)(x - b/2) + h.

Is any of this right? I know my college algebra is a bit rusty.

Yes, I get:

$$y=\frac{2h}{b}x$$

and:

$$y=-\frac{2h}{b}x+2h$$

This second line is equivalent to what you gave.

Now, the region over which we are integration is an isosceles triangle, with its base along the positive $x$-axis...do you think we should use vertical or horizontal strips for simplicity of the computation of iterated double integrals over the region?

 

[harpazo]

Yes, I get:

$$y=\frac{2h}{b}x$$

and:

$$y=-\frac{2h}{b}x+2h$$

This second line is equivalent to what you gave.

Now, the region over which we are integration is an isosceles triangle, with its base along the positive $x$-axis...do you think we should use vertical or horizontal strips for simplicity of the computation of iterated double integrals over the region?

It is my guess that dxdy is the better choice.

The outer limits are 0 to (b/2).

The inner limits are (2h/b)x to (-2h/b)x + 2h.

The inner limits complicate the double integrals not only when searching for the mass but also for the center of mass.

Tell me something, the textbook examples use regular coordinates for points. In the question section, the points given forming our region are far more complicated. What is the purpose of the authors of the calculus 3 textbook to do this? The practice problems are far easier.

 

[MarkFL]

It is my guess that dxdy is the better choice.

I agree...we want to use horizontal strips, because we can then use one integral rather than two.

The outer limits are 0 to (b/2).

The outer limits, in terms of $y$ will go from $y=0$ to $y=h$.

The inner limits are (2h/b)x to (-2h/b)x + 2h.

The inner limits will be the two lines we found, but we need them to be the $x$-values, so we need to solve the two lines for $x$, where the first line we found will give the lower limit (the "left" $x$), and the second line we found will give the upper limit (the "right" $x$).

So, can you now set up the integrals for mass, and center of mass?

The inner limits complicate the double integrals not only when searching for the mass but also for the center of mass.

Tell me something, the textbook examples use regular coordinates for points. In the question section, the points given forming our region are far more complicated. What is the purpose of the authors of the calculus 3 textbook to do this? The practice problems are far easier.

We are expected to be able to generalize from the theorems/examples presented, to be able to use parameters in place of numbers, etc. :D

 

[harpazo]

I agree...we want to use horizontal strips, because we can then use one integral rather than two.
The outer limits, in terms of $y$ will go from $y=0$ to $y=h$.
The inner limits will be the two lines we found, but we need them to be the $x$-values, so we need to solve the two lines for $x$, where the first line we found will give the lower limit (the "left" $x$), and the second line we found will give the upper limit (the "right" $x$).

So, can you now set up the integrals for mass, and center of mass?
We are expected to be able to generalize from the theorems/examples presented, to be able to use parameters in place of numbers, etc. :D

Questions:

1. How did you find the y limits from 0 to h?

2. Solving each equation we found for x will make things messy, right?

3. This is a tough problem no matter how much we try to simplify?

4. Just curious, what would be the set up for dydx?

 

[MarkFL]

Questions:

1. How did you find the y limits from 0 to h?

Let's examine the region over which we are integrating:

\begin{tikzpicture}
[>=stealth, xscale=1, yscale=1, font=\large]
\draw[->] (-0.5,0) -- (8.5,0) node

{$x$};
\draw[->] (0,-0.5) -- (0,6.5) node[above] {$y$};
\draw (6,.1) -- (6,-.1) node[below] {$b$};
\draw (.1,4) -- (-.1,4) node

{$h$};
\draw[blue,thick] (0,0) -- (3,4);
\draw[blue,thick] (3,4) -- (6,0);
\draw[blue,thick] (6,0) -- (0,0);
\node[blue] at (0.75,0.25) {\large $(0,0)$};
\node[blue] at (4,4) {\large $\left(\dfrac{b}{2},h\right)$};
\node[blue] at (6.5,0.25) {\large $(b,0)$};
\end{tikzpicture}

Do you see now how, for the points within the region the $y$-coordinates vary from $0$ to $h$?

2. Solving each equation we found for x will make things messy, right?

Not overly so, in my opinion. :D

3. This is a tough problem no matter how much we try to simplify?

This one is pretty simple actually, we have a simple geometry for the bounded region, and a constant density. :D

4. Just curious, what would be the set up for dydx?

Well, we would need two integrals, because the upper bound for $y$ changes midway. See if you can set up the integrals. ;)​

 

[harpazo]

Thanks, Mark. I see from your diagram that dydx would certainly complicate the problem. I will work on this later.

 

[MarkFL]

Let's see if we can compute the mass using polar coordinates. :D

We see that for $r$, this will vary from $r=0$ to the line given in Cartesian coordinates by:

$$y=-\frac{2h}{b}x+2h$$

Convert to polar coordinates:

$$r\sin(\theta)=-\frac{2h}{b}r\cos(\theta)+2h$$

Solve for $r$:

$$r=\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}$$

We see that $\theta$ ranges from $\theta=0$ to $$\theta=\arctan\left(\frac{2h}{b}\right)$$

And so, for mass $m$, we have:

$$m=k\int_0^{\arctan\left(\frac{2h}{b}\right)} \int_0^{\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}}r\,dr\,d\theta$$

$$m=\frac{k}{2}\int_0^{\arctan\left(\frac{2h}{b}\right)} \left(\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}\right)^2\,d\theta$$

Using a linear combination identity on the denominator of the integrand, there results:

$$m=\frac{2b^2h^2k}{b^2+4h^2}\int_0^{\arctan\left(\frac{2h}{b}\right)} \frac{1}{\sin^2\left(\theta+\arctan\left(\frac{2h}{b}\right)\right)}\,d\theta$$

Let:

$$\beta=\theta+\arctan\left(\frac{2h}{b}\right)\implies d\beta=d\theta$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\int_{\arctan\left(\frac{2h}{b}\right)}^{2\arctan\left(\frac{2h}{b}\right)} \csc^2(\beta)\,d\beta$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\cot\left(\arctan\left(\frac{2h}{b}\right)\right)-\cot\left(2\arctan\left(\frac{2h}{b}\right)\right)\right)$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\frac{b}{2h}-\frac{b^2-4h^2}{4bh}\right)$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\frac{b^2+4h^2}{4bh}\right)$$

$$m=\frac{bhk}{2}$$

This is the product of area of the triangular region and the constant density, which is what we should expect. :D

 

[harpazo]

I really like calculus.

 

[harpazo]

Let's see if we can compute the mass using polar coordinates. :D

We see that for $r$, this will vary from $r=0$ to the line given in Cartesian coordinates by:

$$y=-\frac{2h}{b}x+2h$$

Convert to polar coordinates:

$$r\sin(\theta)=-\frac{2h}{b}r\cos(\theta)+2h$$

Solve for $r$:

$$r=\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}$$

We see that $\theta$ ranges from $\theta=0$ to $$\theta=\arctan\left(\frac{2h}{b}\right)$$

And so, for mass $m$, we have:

$$m=k\int_0^{\arctan\left(\frac{2h}{b}\right)} \int_0^{\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}}r\,dr\,d\theta$$

$$m=\frac{k}{2}\int_0^{\arctan\left(\frac{2h}{b}\right)} \left(\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}\right)^2\,d\theta$$

Using a linear combination identity on the denominator of the integrand, there results:

$$m=\frac{2b^2h^2k}{b^2+4h^2}\int_0^{\arctan\left(\frac{2h}{b}\right)} \frac{1}{\sin^2\left(\theta+\arctan\left(\frac{2h}{b}\right)\right)}\,d\theta$$

Let:

$$\beta=\theta+\arctan\left(\frac{2h}{b}\right)\implies d\beta=d\theta$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\int_{\arctan\left(\frac{2h}{b}\right)}^{2\arctan\left(\frac{2h}{b}\right)} \csc^2(\beta)\,d\beta$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\cot\left(\arctan\left(\frac{2h}{b}\right)\right)-\cot\left(2\arctan\left(\frac{2h}{b}\right)\right)\right)$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\frac{b}{2h}-\frac{b^2-4h^2}{4bh}\right)$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\frac{b^2+4h^2}{4bh}\right)$$

$$m=\frac{bhk}{2}$$

This is the product of area of the triangular region and the constant density, which is what we should expect. :D

Is it harder to integrate with using polar?

 

[harpazo]

Let's see if we can compute the mass using polar coordinates. :D

We see that for $r$, this will vary from $r=0$ to the line given in Cartesian coordinates by:

$$y=-\frac{2h}{b}x+2h$$

Convert to polar coordinates:

$$r\sin(\theta)=-\frac{2h}{b}r\cos(\theta)+2h$$

Solve for $r$:

$$r=\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}$$

We see that $\theta$ ranges from $\theta=0$ to $$\theta=\arctan\left(\frac{2h}{b}\right)$$

And so, for mass $m$, we have:

$$m=k\int_0^{\arctan\left(\frac{2h}{b}\right)} \int_0^{\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}}r\,dr\,d\theta$$

$$m=\frac{k}{2}\int_0^{\arctan\left(\frac{2h}{b}\right)} \left(\frac{2bh}{b\sin(\theta)+2h\cos(\theta)}\right)^2\,d\theta$$

Using a linear combination identity on the denominator of the integrand, there results:

$$m=\frac{2b^2h^2k}{b^2+4h^2}\int_0^{\arctan\left(\frac{2h}{b}\right)} \frac{1}{\sin^2\left(\theta+\arctan\left(\frac{2h}{b}\right)\right)}\,d\theta$$

Let:

$$\beta=\theta+\arctan\left(\frac{2h}{b}\right)\implies d\beta=d\theta$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\int_{\arctan\left(\frac{2h}{b}\right)}^{2\arctan\left(\frac{2h}{b}\right)} \csc^2(\beta)\,d\beta$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\cot\left(\arctan\left(\frac{2h}{b}\right)\right)-\cot\left(2\arctan\left(\frac{2h}{b}\right)\right)\right)$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\frac{b}{2h}-\frac{b^2-4h^2}{4bh}\right)$$

$$m=\frac{2b^2h^2k}{b^2+4h^2}\left(\frac{b^2+4h^2}{4bh}\right)$$

$$m=\frac{bhk}{2}$$

This is the product of area of the triangular region and the constant density, which is what we should expect. :D

I thank you for this great reply. However, I want to solve this without using polar.

Are the inner limits (2hx)/b to (-2hx)/b + 2h or vice versa?

 

[MarkFL]

I thank you for this great reply. However, I want to solve this without using polar.

Yes, I was leaving the Cartesian solution for you to continue...I just wanted to demonstrate the polar method for general information.

Are the inner limits (2hx)/b to (-2hx)/b + 2h or vice versa?

Neither...you want to take the linear equations we found and solve them for $x$, as we wish to integrate from the left $x$ to the right $x$. :D

 

[MarkFL]

Is it harder to integrate with using polar?

I suspect polar was a bit more difficult in this case. :D

 

[harpazo]

Yes, I was leaving the Cartesian solution for you to continue...I just wanted to demonstrate the polar method for general information.
Neither...you want to take the linear equations we found and solve them for $x$, as we wish to integrate from the left $x$ to the right $x$. :D

Your work in terms of polar is super. This is beyond me. I continue to struggle with this question. I will solve the equations you found for x and integrate dxdy.

 

[harpazo]

Solve y = (2hx)/b for x.

by = 2hx

by/2h = x

Is this the lower limit of integration for the inner integral?

 

[MarkFL]

Solve y = (2hx)/b for x.

by = 2hx

by/2h = x

Is this the lower limit of integration for the inner integral?

Yes. (Yes)

 

[harpazo]

Solve y = (-2hx)/b + 2h for x.

y - 2h = (-2hx)/b

b(y - 2h) = -2hx

(by - 2h) = -2hx

(by - 2h)/(-2h) = x

(by/-2h) + 1 = x

Is this the upper limit of integration for the outer integral?

 

[harpazo]

Mark,

I got the wrong answer for mass. I have been playing with this question for 2 days. This is it for me. Can you please do this problem step by step without polar?

My answer for mass is k(bh + 2h)/2. Of course, this is terribly wrong. I can possibly solve questions like this one using regular number for points in the form (x, y). The b and h throw me in for a loop. I give up!

 

[MarkFL]

Solve y = (-2hx)/b + 2h for x.

y - 2h = (-2hx)/b

b(y - 2h) = -2hx

(by - 2h) = -2hx

You've only distributed the b to the first term within the parentheses on the LHS. Personally, I wouldn't distribute:

$$x=\frac{b(2h-y)}{2h}$$

And this will be the upper limit on the inner integral. :D

 

[MarkFL]

Mark,

I got the wrong answer for mass. I have been playing with this question for 2 days. This is it for me. Can you please do this problem step by step without polar?

My answer for mass is k(bh + 2h)/2. Of course, this is terribly wrong. I can possibly solve questions like this one using regular number for points in the form (x, y). The b and h throw me in for a loop. I give up!

Okay, using Cartesian coordinates, we have:

$$m=k\int_0^h\int_{\frac{by}{2h}}^{\frac{b(2h-y)}{2h}}\,dx\,dy$$

Evaluate the inner integral:

$$m=k\int_0^h\left(\frac{b(2h-y)}{2h}-\frac{by}{2h}\right)\,dy$$

Simplify the integrand:

$$m=\frac{k}{2h}\int_0^h 2bh-2by\,dy$$

Apply the FTOC:

$$m=\frac{k}{2h}\left[2bhy-by^2\right]_0^h=\frac{k}{2h}\left(2bh^2-bh^2\right)=\frac{kbh^2}{2h}=\frac{bhk}{2}\quad\checkmark$$ :D

 

[harpazo]

Okay, using Cartesian coordinates, we have:

$$m=k\int_0^h\int_{\frac{by}{2h}}^{\frac{b(2h-y)}{2h}}\,dx\,dy$$

Evaluate the inner integral:

$$m=k\int_0^h\left(\frac{b(2h-y)}{2h}-\frac{by}{2h}\right)\,dy$$

Simplify the integrand:

$$m=\frac{k}{2h}\int_0^h 2bh-2by\,dy$$

Apply the FTOC:

$$m=\frac{k}{2h}\left[2bhy-by^2\right]_0^h=\frac{k}{2h}\left(2bh^2-bh^2\right)=\frac{kbh^2}{2h}=\frac{bhk}{2}\quad\checkmark$$ :D

Questions:

1. What FTOC?

2. Do I use the same upper and lower limits to set up and find the center of mass?

 

[MarkFL]

Questions:

1. What FTOC?

The anti-derivative form:

$$\int_a^b f(x)\,dx=F(b)-F(a)$$ where $$\d{F}{x}=f(x)$$

2. Do I use the same upper and lower limits to set up and find the center of mass?

Yes, they describe the same region. :D

 

[MarkFL]

I think what I would do to find the center of mass, is to use the left-right symmetry of the region, and orient the coordinate axes such that the coordinates of the vertices are:

$$\left(-\frac{b}{2},0\right),\,(0,h),\,\left(\frac{b}{2},0\right)$$

And so we have:

$$\overline{x}=\frac{2}{bh}\int_0^h \int_{-\frac{b}{2h}(h-y)}^{\frac{b}{2h}(h-y)}x\,dx\,dy$$

Odd-function rule:

$$\overline{x}=0$$

$$\overline{y}=\frac{2}{bh}\int_0^h y\int_{-\frac{b}{2h}(h-y)}^{\frac{b}{2h}(h-y)}\,dx\,dy$$

$$\overline{y}=\frac{2}{bh}\int_0^h y\frac{b}{h}(h-y)\,dy=\frac{2}{h}\int_0^h y-\frac{1}{h}y^2\,dy=\frac{h}{3}$$

And so this means for the original coordinate system, we would have:

$$\left(\overline{x},\overline{y}\right)=\left(\frac{b}{2},\frac{h}{3}\right)$$ :D

 

[harpazo]

Good job. Great data for my study notes. It's time to move on to other questions. This is not too bad if the region given had actual numbers instead of letters. If the region was (0,0), (2,2) and (3,0), I would be able to find the mass and center of mass easily. The variables b and h created confusion for me. Thanks again. I posted two new questions from the same section with a slight twist.