Find the bounds after changing the variables in a double integral

[Leo Liu]

Homework Statement

.

Relevant Equations

.

The answer calculates the integral with $du$ before $dv$ as shown below.

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.

 

[BvU]

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 $v$

$\$

 

[Leo Liu]

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.

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?

 

[BvU]

You can evaluate and check for yourself ...

$\$

 

[Leo Liu]

You can evaluate and check for yourself ...

$\$

I think I got it right. I appreciate your help.

 

[BvU]

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)