How do I set up double integrals for different orders of integration?

[harpazo]

Set up an integral for both orders of integration.
DO NOT EVALUATE THE INTEGRAL.

Let S S = double integrals

Let R = region

S S xe^(y) dA

R: triangle bounded by y = 4 - x, y = 0, x = 0

I can graph the region but have no idea how to proceed from there. I need solution steps.

 

[MarkFL]

If we use horizontal strips, then we have:

\(\displaystyle I=\int_0^4 e^y\int_0^{4-y} x\,dx\,dy\)

And if we use vertical strips, we have:

\(\displaystyle I=\int_0^4 x\int_0^{4-x}e^y\,dy\,dx\)

 

[harpazo]

If we use horizontal strips, then we have:

\(\displaystyle I=\int_0^4 e^y\int_0^{4-y} x\,dx\,dy\)

And if we use vertical strips, we have:

\(\displaystyle I=\int_0^4 x\int_0^{4-x}e^y\,dy\,dx\)

How do you know which limits of integration to apply to dxdy as oppossed to dydx? This is my biggest problem.

 

[MarkFL]

How do you know which limits of integration to apply to dxdy as oppossed to dydx? This is my biggest problem.

That's where the sketch of the region $D$ comes in...for example when using horizontal strips, we see they are bounded on the left by the line $x=0$ and on the right by $x=4-y$. We then observe that the strips run from $y=0$ to $y=4$.

If we use vertical strips, then we see they are bounded on the bottom by the line $y=0$ to the line $y=4-x$, and that these strips run from $x=0$ to $x=4$. :D

 

[harpazo]

That's where the sketch of the region $D$ comes in...for example when using horizontal strips, we see they are bounded on the left by the line $x=0$ and on the right by $x=4-y$. We then observe that the strips run from $y=0$ to $y=4$.

If we use vertical strips, then we see they are bounded on the bottom by the line $y=0$ to the line $y=4-x$, and that these strips run from $x=0$ to $x=4$. :D

I am having a hard time deciding how to set up dxdy or dydx based on graphs of general regions.