Determining the limits of a double integral over a given region

[wind moss]

Hello, can you help me find the limits of these two integrals:

[wind moss]

$\int \int \frac{dxdy}{(2x+3y)^2}, R:(x=0, x=1, y=1, y=2)$

[hasty cairn]

The Mysterious Stranger

[wind moss]

$\int \int \frac{dxdy}{(x+y)^3}, G:(x\geq1 , y\geq1, x+y\leq 3)$

[hasty cairn]

The Mysterious Stranger

[dusky dome]

So the dx is first right?

This means the first integral would have x-bounds

x = 0
x = 1
As the lower and upper bound respectively for the first integral. The second integral would be the same but with y-values because the second differential is dy.

[dusky dome]

For the second one, find where are all the three inequalities hold. If you let x be fixed, what would y range to?

x + y ≤ 3

You can know what y would range to [1, y] by solving that inequality

This creates a triangular region

This explanation is not the best but lmk what’s confusing

[wind moss]

That is, the first integral is of the form:

$\int_{y=1}^{y=2}dy \int_{x=0}^{x=1} \frac{dx}{(2x+3y)^2}$

[hasty cairn]

The Mysterious Stranger

[wind moss]

the second one:

$\int_{x=1}^{x=2}dx \int_{y=1}^{3-x} \frac{dy}{(x+y)^3}$

[hasty cairn]

The Mysterious Stranger

[wind moss]

Is that correct?

[dusky dome]

Should be yeah