the-volume-of-the-region-between-the-planes-x-y-2z-2-and-2x-y-z-4-in-the-first-octant-is-

Question Number 210723 by universe last updated on 17/Aug/24

$$\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{between}\:\mathrm{the}\: \\ $$$$\mathrm{planes}\:\mathrm{x}+\mathrm{y}+\mathrm{2z}=\mathrm{2}\:\mathrm{and}\:\mathrm{2x}+\mathrm{y}+\mathrm{z}\:=\:\mathrm{4}\:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{octant}\:\mathrm{is} \\ $$

Answered by mr W last updated on 17/Aug/24

$${V}=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{2}×\mathrm{4}×\mathrm{4}−\mathrm{2}×\mathrm{2}×\mathrm{1}\right)=\frac{\mathrm{14}}{\mathrm{3}} \\ $$

Commented by universe last updated on 17/Aug/24

$${sir}\:{can}\:{u}\:{explain}\:{your}\:{method}\:{little}\:{bit} \\ $$$${and}\:{how}\:{tw}\:{solve}\:{by}\:{integration}\:{method} \\ $$

Commented by mr W last updated on 17/Aug/24

Commented by mr W last updated on 17/Aug/24

the eqn. of blue plane is (x/a)+(y/b)+(z/c)=1 the volume under the plane is V_1 =((ab)/2)×(c/3)=((abc)/6) similarly the volume under the red plane (x/p)+(y/q)+(z/r)=1 is V_2 =((pqr)/6) the volume between blue plane and red plane is V=V_1 −V_2 =(1/6)(abc−pqr). in this example the red plane is x+y+2z=2, i.e. (x/2)+(y/2)+z=1 and the blue plane is 2x+y+z=4, i.e. (x/2)+(y/4)+(z/4)=1. therefore V=(1/6)(2×4×4−2×2×1)=((14)/3)

$${the}\:{eqn}.\:{of}\:{blue}\:{plane}\:{is} \\ $$$$\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}}+\frac{\boldsymbol{{y}}}{\boldsymbol{{b}}}+\frac{\boldsymbol{{z}}}{\boldsymbol{{c}}}=\mathrm{1} \\ $$$${the}\:{volume}\:{under}\:{the}\:{plane}\:{is} \\ $$$${V}_{\mathrm{1}} =\frac{{ab}}{\mathrm{2}}×\frac{{c}}{\mathrm{3}}=\frac{{abc}}{\mathrm{6}} \\ $$$${similarly}\:{the}\:{volume}\:{under}\:{the} \\ $$$${red}\:{plane}\:\frac{\boldsymbol{{x}}}{\boldsymbol{{p}}}+\frac{\boldsymbol{{y}}}{\boldsymbol{{q}}}+\frac{\boldsymbol{{z}}}{\boldsymbol{{r}}}=\mathrm{1}\:{is} \\ $$$${V}_{\mathrm{2}} =\frac{{pqr}}{\mathrm{6}} \\ $$$${the}\:{volume}\:{between}\:{blue}\:{plane}\:{and} \\ $$$${red}\:{plane}\:{is} \\ $$$${V}={V}_{\mathrm{1}} −{V}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\left({abc}−{pqr}\right). \\ $$$${in}\:{this}\:{example}\:{the}\:{red}\:{plane}\:{is} \\ $$$${x}+{y}+\mathrm{2}{z}=\mathrm{2},\:{i}.{e}.\:\frac{{x}}{\mathrm{2}}+\frac{{y}}{\mathrm{2}}+{z}=\mathrm{1} \\ $$$${and}\:{the}\:{blue}\:{plane}\:{is}\: \\ $$$$\mathrm{2}{x}+{y}+{z}=\mathrm{4},\:{i}.{e}.\:\frac{{x}}{\mathrm{2}}+\frac{{y}}{\mathrm{4}}+\frac{{z}}{\mathrm{4}}=\mathrm{1}. \\ $$$${therefore} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{2}×\mathrm{4}×\mathrm{4}−\mathrm{2}×\mathrm{2}×\mathrm{1}\right)=\frac{\mathrm{14}}{\mathrm{3}} \\ $$

Commented by mr W last updated on 18/Aug/24