Question

How would you find the volume bounded by the coordinate planes and by the plane 3x + 2y + 2z = 6?

Answer 1

`V=int_(x=0)^(x=2) int_(y=0)^(y=3-3/2x) int_(z=0)^(z=3-3/2x-y)dzdydx=3`

Explanation:

`3x+2y+2z=6`

By solving for `z`,

`z=3-3/2x-y`

By setting `z=0` and solving for `y`,

`y=3-3/2x`

By setting `y=0` and solving for `x`,

`x=2`

Now, the volume of the solid can be expressed as:

`V=int_(x=0)^(x=2) int_(y=0)^(y=3-3/2x) int_(z=0)^(z=3-3/2x-y)dzdydx`

`=int_(x=0)^(x=2) int_(y=0)^(y=3-3/2x) [z]_(z=0)^(z=3-3/2x-y)dydx`

`=int_(x=0)^(x=2) int_(y=0)^(y=3-3/2x) (3-3/2x-y)dydx`

`=int_(x=0)^(x=2) [(3-3/2x)y-y^2/2]_(y=0)^(y=3-3/2x) dx`

`=int_(x=0)^(x=2) (3-3/2x)^2/2 dx`

`=[(3-3/2x)^3/(cancel(2)cdot3cdot(-3/cancel(2)))]_0^2=[-(3-3/2x)^3/9]_0^2`

`=0-(-3)=3`

I hope that this was clear.