i need to find the volume bounded by

x + z = 1 and y + 2z=2 in the positive quadrant

 

ive worked out the co-ordinates which are:

x(1,0,0) y(0,2,0) Z(0,0,1)

 

how do i find out the limits to put for the three integrals.

i think the first integral is going to be:

0 to (1-z)

2nd integral: 0 to (2-2z)/2

 

but im not sure about these two either can some clarify for me if im going about this the right way.

 

im not sure about the third one.

can any one help.

thanx

you seem to have problems with integration I suggest talking to a TA about it because even though forums give you the fast answer you should really learn to do it properly.

 

Anyhoo: work out the relational eq'n b/w x and y. This should be easy since you already have the coordinates. Draw it out if your having problems.

Then

your x integral should be values in this case 0:1

the y integral has one limit dependent on x

and teh z integral will have both limits dependent on x or one on x and one on y.

i think i should explain my situation, i have come back to maths after some time, and im in a situation where i cannot ask any one as this is a course is studied from home. no one i know, knows integration, therefore i was getting help from u guys, as i then know if im heading the right way.

i hope u dont mind, and this is ok with u.

I haven't started 3space integrals, so I can't help you directly, but this site: http://www.sosmath.com/CBB/index.php helps people with dilemnas like yours.

 

Check it out!

oh ok...well then, you have 2 eq'n, an xz and an yz equation and you found your coordinates...you should be able to draw out a 3D image as I suggested above to find an xy relation because your volume has a y dependent on x. (I hope you know how to draw 3D-diagrams cuz theyare very essential in visualizing integration).

Next you'll have to find teh plane equation that results in the positive quadrant that relates x,y,z

 

then your integral will

 

int(int(int 1 dz)dy)dx...

 

your x limits are 0:1

your y limits are 0:i'll let you figure out dependent on x

your z limits are 0:the plane equation z dependent on x and y.

 

OR you can do int(int(int 1 dy)dz)dx

your x limits are 0:1

your z limits are 0:one of the two given equations

your y lmits are 0:the plane equation rearranged for y dependent

on z,x

yep i get the diagram,

 

to find the equation of the plane, is it possible to let the two equations equal each other, ie,

1-x=1-(y/2)

simplified to z = 2-y-2x

 

as this connects x and y.

no that just gives you the x and y relation...

to find the equation of the plane have you learned about normals?? specifically this equation

Ax+By+Cz+D = 0? thats what you should be looking for

okey dokey i have worked out the limits according to me i get:

 

1st integral: 0 : 1

2nd integral: 0 : 2-2x

3rd integral: 0 : (2-y-2x)/2

 

i have worked this out by dz dy dx and i get the volume to be 1.

can some one verify that 4 me.

thanx

 

oh by the way Neurocomp2003 i am really appreciating ur help!!! thanx alot