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# How to find the region of integration, dA?

0
1
839
3
+42

and so on....

What's really annoying me is that the answers gloss over the mechanical integrating part (did the same in the lecture notes too) so I have no idea what region dA represents... I can obtain the function no problem but how do i determine what region S1 is , region S2 etc (btw it's probably a typo inbetween point (9) and (10) , think it meant to say the part of the surface with z=0)

Nov 8, 2015

#2
+28179
+10

I think the first part is as follows (but I don't guarantee it!!):

For the second part the kernel of the integration seems to be identically zero.

Nov 8, 2015

#1
+42
0

Just to make what I'm asking as clear as possible:

The entire surface S is bounded by:   y=0  ,   z =0   ,  4x + y + 2z = 4    above the y-z plane

Part of this surface, $$S_1$$, is the surface along y=0.

Find $$\int\int_{S_1} y^2-x dA$$

What are the bounds of integration I should be using here? What is dA meant to be (Is it dx dy or is it dx dz or something else) ?

Nov 8, 2015
#2
+28179
+10

I think the first part is as follows (but I don't guarantee it!!):

For the second part the kernel of the integration seems to be identically zero.

Alan Nov 8, 2015
#3
+42
+5

I drew a quick representation of what I thought "above the y - z plane " means. I interpreted it as "above the y plane" means all values of y > 0 and "above the z plane" means all values of z above zero. I completely agree with the rest of your working but this is the only source of confusion that I still have left. Your way seems correct as it returns a lower bound for x whilst my interpretation lets x go down to negative infinity.

Edit:

Did a google search and saw that the x-y plane is where x = x , y = y , z = 0. So in this case the y-z plane would be where x = 0 , y = y , z = z  and "above" means it's on the positive side. My confusion is cleared and I agree with your answer Alan, many thanks =)

Nov 8, 2015
edited by Maximillian  Nov 8, 2015