Question

V - domain in 1st quadrant limited with flat : x/2 + y/2 + z = 1

Answer 1

I need to find V or volume

Answer 2

which method do i use here?

Answer 3

multi-integral

Answer 4

in this case a triple integral

Answer 5

what are the boundaries of your curve?

Answer 6

i don't know how to find the boundaries

Answer 7

ok in order to integrate something, you must first know its boundaries. do you know how to compute a single integral?

Answer 8

yes..when i have it's boundaries

Answer 9

find \(\int x\;dx\)

Answer 10

here i tried using jacobian and changing x/2 , y/2 and z to u , v and w but i am not quite sure if i got the calculation right for the matrix

Answer 11

you do not need a Jacobian here; only the boundaries.

Answer 12

but how do i get them?
what am i supposed to do with the given equation?

Answer 13

integral of xdx is x^2 / 2

Answer 14

recall that \(\frac{1}{2}x+\frac{1}{2}y+z=1\) is a plane

Answer 15

when you integrate some curve of one independent variable, you add a set of rectangles

Answer 16

so do i use the y = ax + b ?

Answer 17

no, you must find where the plane \(\frac{1}{2}x+\frac{1}{2}y+z=1\) intersects the xy-plane, yz-plane and zx-plane.

Answer 18

okay, how do i do that?

Answer 19

ok, let us first observe the order of the differentials (dx dy dz)

Answer 20

the limits of the first integral are 0 to the boundary of the given plane and the yz-plane

Answer 21

i.e. where they intersect

Answer 22

so where do they intersect?
do i use any values for x and y or is there a way to calculate this?

Answer 23

what is the x coordinate of any point on the yz-plane?

Answer 24

hint:
the situation can be represented by this drawing: