Find the bounds after changing the variables in a double integral

  • Thread starter Leo Liu
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  • #1
Leo Liu
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Homework Statement:
.
Relevant Equations:
.
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.
 

Answers and Replies

  • #2
BvU
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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##

##\ ##
 
  • #3
Leo Liu
345
145
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
BvU
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You can evaluate and check for yourself ...

##\ ##
 
  • #5
Leo Liu
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You can evaluate and check for yourself ...

##\ ##
I think I got it right. I appreciate your help.
1616364743893.png
 
  • #6
BvU
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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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