Vactor Calculus: Green's Theorem Problem Clarification

pope4

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Hi! I was going through a question from my textbook and I think I'm missing some conceptual understanding of the topic because I'm having a hard time grasping the provided solution.

Question:

1754343995642.png


Solution:

1754344023000.png
1754344037748.png


I understand the simplification after applying Green's Theorem, but I'd like to double check if I'm understanding their reasoning for getting rid of the double integral of 2y correctly. The way I imagine it starts by assuming that 3+2y is a function in the z-axis (since we're integrating with respect to x and y), giving:

[math]z = 3 +2y[/math]
Then, since we're splitting the intergal and only looking at 2y, we ignore the 3 for now. The integration domain (i.e. the trapezoid) is symmetric across the x-axis, and since we're calculating the volume from [imath]z=2y[/imath] to the xy-plane with [imath]z = |2y|[/imath] being symmetric across the zx-plane, the first integral will be 0, meaning the second one will also yield 0. In other words, the volumes under [imath]z = 2y[/imath] on either side of the z-axis will be equal in magnitude but opposite in sign, so they cancel out. Is that correct? If so, how would I go about recognizing this type of pattern in the future? I only realized this after seeing the solution and drawing out the surfaces, and this time the equations were quite nice. However, assuming this doesn't always work since not all equations will be symmetric in this way, is this type of analysis simply a habit I should just get used to, even with harder equations?

Thank you so much in advance!
 
Once you use Green's theorem and get the solution say this:

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy\)

You have three options to evaluate it.

\(\displaystyle \textcolor{blue}{\bold{First.}}\)

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy = \bold{average \ value} \times \bold{trapezoid \ area}\)

Let \(\displaystyle f(x,y) = 3 + 2y\), then

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy = \frac{f(0,-2) + f(1,-1) + f(1,1) + f(0,2)}{4} \times \frac{1}{2}(2 + 4)(1)\)

\(\displaystyle \textcolor{blue}{\bold{Second.}}\)

but I'd like to double check if I'm understanding their reasoning for getting rid of the double integral of 2y correctly.
Here is the answer to your question. \(\displaystyle 2y\) vanishes because the average value of \(\displaystyle y\) is \(\displaystyle \frac{2 - 2}{2} = 0\).

Here the second way to solve the integral.

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy = \bigg[3 + 2\bold{(average \ value \ of \ y)}\bigg] \times \bold{trapezoid \ area}\)

Or

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy = \bigg[3 + 2(0)\bigg] \times \frac{1}{2}(2 + 4)(1)\)

\(\displaystyle \textcolor{blue}{\bold{Third.}}\)

This method is the one that \(\displaystyle 99\%\) of students use. Just evaluate the integral normally.

The \(\displaystyle y\) variable goes from the bottom line \(\displaystyle y = x - 2\) to the upper line \(\displaystyle y = 2 - x\) while the variable \(\displaystyle x\) goes from \(\displaystyle 0\) to \(\displaystyle 1\).

\(\displaystyle \iint\limits_T 3 + 2 y \ dx \ dy = \int_{0}^{1}\int_{x - 2}^{2 - x} 3+2y \ dy \ dx\)
 
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In the \(\displaystyle \textcolor{blue}{\bold{second}}\) method, we calculated the average value of \(\displaystyle y\) as:

\(\displaystyle y_{\text{av}} = \frac{\text{lowest point} + \text{highest point}}{2} = \frac{-2 + 2}{2} = \frac{0}{2} = 0\)

We could do that because of the symmetry of the region around the \(\displaystyle x\)-axis. But if there is no symmetry, we have to do some extra work.

Suppose that we have this region:

line_integral.png

Here it is wrong to say that the average value of \(\displaystyle y\) is \(\displaystyle \frac{-2 + 2}{2} = 0\) because the region is not symmetric with the \(\displaystyle x\)-axis.

This is a great example to show you what to do when the region is not symmetric. If you are interested to know how to solve this non-symmetric region, let me know!

;)😉
 
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In the \(\displaystyle \textcolor{blue}{\bold{second}}\) method, we calculated the average value of \(\displaystyle y\) as:

\(\displaystyle y_{\text{av}} = \frac{\text{lowest point} + \text{highest point}}{2} = \frac{-2 + 2}{2} = \frac{0}{2} = 0\)

We could do that because of the symmetry of the region around the \(\displaystyle x\)-axis. But if there is no symmetry, we have to do some extra work.

Suppose that we have this region:

View attachment 39644

Here it is wrong to say that the average value of \(\displaystyle y\) is \(\displaystyle \frac{-2 + 2}{2} = 0\) because the region is not symmetric with the \(\displaystyle x\)-axis.

This is a great example to show you what to do when the region is not symmetric. If you are interested to know how to solve this non-symmetric region, let me know!

;)😉

That makes sense, thank you so much for your help!

I'd love to know how to solve the asymmetric problems as well, if that's alright! Would you just split it into 2 right triangles and sum the separate integrals over those 2 regions?
 
That makes sense, thank you so much for your help!
You're welcome.

I'd love to know how to solve the asymmetric problems as well, if that's alright!
Of course.

Would you just split it into 2 right triangles and sum the separate integrals over those 2 regions?
Not two triangles but rather \(\displaystyle 1\) trapezoid (like our previous trapezoid) and \(\displaystyle 1\) triangle. So, yes, we split it into two regions. I will show you the sketch.

line_integral_2.png
So our integral will be:

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy = \iint\limits_{R_1} (3 + 2y) \ dx \ dy + \iint\limits_{R_2} (3 + 2y) \ dx \ dy\)

You already know how to solve the first region and we also have solved it in three different methods. So we will focus on region \(\displaystyle 2\).

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy\)

\(\displaystyle \textcolor{red}{\bold{First \ Method}}\)

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \textcolor{black}{\bold{average \ value}} \times \textcolor{black}{\bold{triangle \ area}}\)

Let \(\displaystyle f(x,y) = 3 + 2y\)

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \frac{f(1,-1) + f(3,1) + f(1,1)}{3} \times \frac{1}{2}2(2)\)


\(\displaystyle \textcolor{red}{\bold{Third \ Method}}\)

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \int_{1}^{3}\int_{x - 2}^{1} (3 + 2y) \ dy \ dx\)

\(\displaystyle \textcolor{blue}{\bold{Second \ Method}}\)

I reversed the order of method \(\displaystyle 2\) and \(\displaystyle 3\) because if the region is not symmetric method \(\displaystyle 2\) is just the same as method \(\displaystyle 3\) but longer.

Let me show you what I mean.

In method two we want to find the average value of \(\displaystyle y\) or more precisely you want to find the average value of \(\displaystyle (3 + 2y)\).

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \bigg[\textcolor{black}{\bold{average \ value \ of \ }} (3 + 2y)\bigg] \times \textcolor{black}{\bold{triangle \ area}}\)

Or

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \bigg[\frac{1}{\textcolor{black}{\bold{triangle \ area}}}\int_{1}^{3}\int_{x - 2}^{1} (3 + 2y) \ dy \ dx\bigg] \times \textcolor{black}{\bold{triangle \ area}} = \int_{1}^{3}\int_{x - 2}^{1} (3 + 2y) \ dy \ dx\)


For the sake of completeness let me give you the final answer.

\(\displaystyle \iint\limits_{R_1} (3 + 2y) \ dx \ dy = 9\)

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \frac{22}{3}\)

Then,

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy = 9 + \frac{22}{3} = \textcolor{violet}{\frac{49}{3}}\)
 
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Would you just split it into 2 right triangles and sum the separate integrals over those 2 regions?
Of course we can also split the region into \(\displaystyle 2\) right triangles, or anything else we want. But when we have the opportunity to get a symmetrical region we will certainly go for it as it makes the calculations easier!

🥳🙌
 
Beer induced query follows.
Hi! I was going through a question from my textbook and I think I'm missing some conceptual understanding of the topic because I'm having a hard time grasping the provided solution.

Question:

View attachment 39638


Solution:

View attachment 39639
View attachment 39640


I understand the simplification after applying Green's Theorem, but I'd like to double check if I'm understanding their reasoning for getting rid of the double integral of 2y correctly. The way I imagine it starts by assuming that 3+2y is a function in the z-axis (since we're integrating with respect to x and y), giving:

[math]z = 3 +2y[/math]
Then, since we're splitting the intergal and only looking at 2y, we ignore the 3 for now. The integration domain (i.e. the trapezoid) is symmetric across the x-axis, and since we're calculating the volume from [imath]z=2y[/imath] to the xy-plane with [imath]z = |2y|[/imath] being symmetric across the zx-plane, the first integral will be 0, meaning the second one will also yield 0. In other words, the volumes under [imath]z = 2y[/imath] on either side of the z-axis will be equal in magnitude but opposite in sign, so they cancel out. Is that correct? If so, how would I go about recognizing this type of pattern in the future? I only realized this after seeing the solution and drawing out the surfaces, and this time the equations were quite nice. However, assuming this doesn't always work since not all equations will be symmetric in this way, is this type of analysis simply a habit I should just get used to, even with harder equations?

Thank you so much in advance!
What is the ISBN of your textbook?
 
You're welcome.


Of course.


Not two triangles but rather \(\displaystyle 1\) trapezoid (like our previous trapezoid) and \(\displaystyle 1\) triangle. So, yes, we split it into two regions. I will show you the sketch.

View attachment 39645
So our integral will be:

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy = \iint\limits_{R_1} (3 + 2y) \ dx \ dy + \iint\limits_{R_2} (3 + 2y) \ dx \ dy\)

You already know how to solve the first region and we also have solved it in three different methods. So we will focus on region \(\displaystyle 2\).

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy\)

\(\displaystyle \textcolor{red}{\bold{First \ Method}}\)

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \textcolor{black}{\bold{average \ value}} \times \textcolor{black}{\bold{triangle \ area}}\)

Let \(\displaystyle f(x,y) = 3 + 2y\)

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \frac{f(1,-1) + f(3,1) + f(1,1)}{3} \times \frac{1}{2}2(2)\)


\(\displaystyle \textcolor{red}{\bold{Third \ Method}}\)

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \int_{1}^{3}\int_{x - 2}^{1} (3 + 2y) \ dy \ dx\)

\(\displaystyle \textcolor{blue}{\bold{Second \ Method}}\)

I reversed the order of method \(\displaystyle 2\) and \(\displaystyle 3\) because if the region is not symmetric method \(\displaystyle 2\) is just the same as method \(\displaystyle 3\) but longer.

Let me show you what I mean.

In method two we want to find the average value of \(\displaystyle y\) or more precisely you want to find the average value of \(\displaystyle (3 + 2y)\).

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \bigg[\textcolor{black}{\bold{average \ value \ of \ }} (3 + 2y)\bigg] \times \textcolor{black}{\bold{triangle \ area}}\)

Or

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \bigg[\frac{1}{\textcolor{black}{\bold{triangle \ area}}}\int_{1}^{3}\int_{x - 2}^{1} (3 + 2y) \ dy \ dx\bigg] \times \textcolor{black}{\bold{triangle \ area}} = \int_{1}^{3}\int_{x - 2}^{1} (3 + 2y) \ dy \ dx\)


For the sake of completeness let me give you the final answer.

\(\displaystyle \iint\limits_{R_1} (3 + 2y) \ dx \ dy = 9\)

\(\displaystyle \iint\limits_{R_2} (3 + 2y) \ dx \ dy = \frac{22}{3}\)

Then,

\(\displaystyle \iint\limits_T (3 + 2y) \ dx \ dy = 9 + \frac{22}{3} = \textcolor{violet}{\frac{49}{3}}\)

Ohhh cool! Thanks so much, that helps a lot!
 
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