Find the bounds after changing the variables in a double integral

In summary, the conversation discusses the calculation of an integral with du before dv and the decision to compute it in the opposite order with different bounds. The conversation also includes a discussion of the correctness of the bounds and the suggested approach to evaluate and check the integral.
  • #1
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Homework Statement
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Relevant Equations
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1616299276442.png

The answer calculates the integral with ##du## before ##dv## as shown below.
1616299342699.png

However I decided to compute it in the opposite order with different bounds. Here is my work:
According to the definitions, $$\begin{cases} u=x+y\\ v=2x-3y \end{cases}$$
First we need to convert the boundaries in xy into uv. The equation of y-axis is ##v=-3u## and the equation of x-axis is ##v=2u##. This two lines give the lower and upper limits of the inner integral dv. The bounds for u can be found by substituting ##x=0## and ##y=0## into ##2x-3y=4##. The lower bounds appears at the lower vertex of the triangle with the value of ##u=-4/3##. Similarly, the upper bounds is the right vertex at ##u=2##. After putting things together we get
$$\frac 1 5 \int _{-4/3}^{2}\int _{-3u}^{2u}v^2u^2\,dvdu$$
I'd like someone to check if the bounds in this integral are correct. Thanks in advance.
 
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  • #2
Hi,

If you did everything correctly, your integral would yield the same as what the alternative yields. I get a different result, though.

Draw some lines where ##u## is constant and check the bounds of ##v##

##\ ##
 
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  • #3
BvU said:
Hi,

If you did everything correctly, your integral would yield the same as what the alternative yields. I get a different result, though.

Draw some lines where ##u## is constant and check the bounds of ##v##
Thank you! I think I messed up the bounds for v.
123.jpg

The graph above shows that heights of the endpoints of the line segment spanning the height of the triangle at a fixed x value are ##v=4## and ##v=-3u## or ##v=2u##; the choice of the lower bound depends on x. Hence the double integral can be written as

$$\frac 1 5 \int _{-4/3}^{0}\int _{-3u}^{4}v^2u^2\,dvdu+\frac 1 5 \int _{0}^{2}\int _{2u}^{4}v^2u^2\,dvdu$$
Is this correct?
 
  • #4
You can evaluate and check for yourself ...

##\ ##
 
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  • #5
BvU said:
You can evaluate and check for yourself ...

##\ ##
I think I got it right. I appreciate your help.
1616364743893.png
 
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  • #6
Looks ok. (didn't check it, but I see you have to do two integrals -- which is the reason the book solution prefers the other order)
 

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