+0  
 
0
340
3
avatar+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)

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Maximillian  Nov 8, 2015

Best Answer 

 #2
avatar+26329 
+10

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

 

surface integrals

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

Alan  Nov 8, 2015
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3+0 Answers

 #1
avatar+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) ?

Maximillian  Nov 8, 2015
 #2
avatar+26329 
+10
Best Answer

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

 

surface integrals

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

Alan  Nov 8, 2015
 #3
avatar+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 =)

Maximillian  Nov 8, 2015
edited by Maximillian  Nov 8, 2015

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