Double integral of triangle

[pj33]

How can I calculate the
[MATH] \iint_S y \,dx\,dy [/MATH]for a triangle of corners ar (0,0),(1,2),(-1,1)?
I am a bit confused of how to set the limits of the doulble integrals.
Can someone explaine me please?

Thank you in advance

 

[Deleted member 4993]

How can I calculate the
[MATH] \iint_S y \,dx\,dy [/MATH]for a triangle of corners ar (0,0),(1,2),(-1,1)?
I am a bit confused of how to set the limits of the doulble integrals.
Can someone explaine me please?

Thank you in advance

This problem is same as that of the 2nd. problem posted in:

Help with double integral

Hi, can someone help me with the following double integrals. please? Thanks!! 1 - 2 -

www.freemathhelp.com

although from different author and IP address.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[HallsofIvy]

How can I calculate the
[MATH] \iint_S y \,dx\,dy [/MATH]for a triangle of corners ar (0,0),(1,2),(-1,1)?
I am a bit confused of how to set the limits of the doulble integrals.
Can someone explaine me please?

Thank you in advance

First I suggest drawing a rough graph to get an idea what this looks like.

You should see that the line y= -x connects (0, 0) and (-1, 1), that the line y= 2x connects (0, 0) and (1, 2), and that the line y= (x+ 3)/2, above both of the lines, connects (-1, 1) and (1, 2).

Since the "bottom" of the region consists of two different lines, I would do the integral in two parts. For x from -1 to 0, y goes from the lower y= -x to the upper y= (x+ 3)/2. The integral is \(\displaystyle \int_{-1}^0 \int_{-x}^{(x+3)/2} y dy dx\).

For x from 0 to 1, y goes from the lower y= 2x to the upper y= (x+3)/2. The integral is \(\displaystyle \int_0^1\int_{2x}^{(x+3)/2} y dydx\).

The entire integral is \(\displaystyle \int_{-1}^0 \int_{-x}^{(x+3)/2} y dy dx+ \int_0^1\int_{2x}^{(x+3)/2} y dydx\).