- #1

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- Homework Statement
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- Relevant Equations
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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.