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

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Answer 1 · 2017-04-14T01:33:38

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

#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.