Question

How do you find the volume of the solid bounded by the coordinate planes and the plane `8x + 6y + z = 6`?

Answer 1

`3/4`

Explanation:

use double integral (or triple if you like, i'll just do double as triple here is just extra unnecessary formality)

first we need to find the volume in question.

`8x + 6y + z = 6`

it hits the x,y,z axes as follows

`y,z = 0, x = 3/4`

`x,z = 0, y = 1`

`x,y = 0, z = 6`

so we can start with a drawing!!

so it's just a case now of finding the integration limits for this double integral `int int \ z(x,y) dA = int int \ 6 - 8x - 6y \ dx \ dy`

in the x-y plane we have

so either of these is fine

`int_{y=0}^{1} int_{x=0}^{3/4 (1-y)} \ 6 - 8x - 6y \ dx \ dy`

`int_{x=0}^{3/4} int_{y=0}^{1 - 4/3 x} \ 6 - 8x - 6y \ dy \ dx`

Either way, I get answer 3/4