Area of $f(x,y) = \frac{1}{(2x+3y)^2}$

Question

I was asked to find the area of $f(x,y) = \frac{1}{(2x+3y)^2}$ inside the parallelogram defined by the points (5,-1); (8, -3); (6, -1) and (9, -3).

I am stuck trying to find a suitable change of variables in order to integrate inside of a rectangle. I attempted with the trivial change $u=2x$ and $v = 3y$ but it just defines another parallelogram.

What alternative path could I follow?

Thank you very much.

The integrating part is easy, the thing is inside what limits to calculate it. — John, Dec 01 '14 at 17:57

score 0 · Answer 1 · answered Dec 01 '14 at 16:59

0

Let $A = (5,-1), B = (8,-3), C = (6,-1), D = (9,-3)$, then the line $AB$ has equation: $2x+3y = 7$, and the line $CD$ has equation: $2x + 3y = 9$. The line $AC$ has equation: $y = -1$, and the line $BD$ has equation: $y = -3$. We can take $u = 2x+3y$, and $v = 0x + 1y = y$, then you are back to the rectangle.

answered Dec 01 '14 at 16:59

DeepSea

77,651

Great! I was over complicating it, trying to make it work so that I had to use U and V through an operation. I didn't think of leaving a variable free. Thanks a lot, OC-Sansoo. – John Dec 01 '14 at 17:55

Area of $f(x,y) = \frac{1}{(2x+3y)^2}$

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