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Question Number 210723 by universe last updated on 17/Aug/24
the volume of the region between the   planes x+y+2z=2 and 2x+y+z = 4 in   the first octant is
thevolumeoftheregionbetweentheplanesx+y+2z=2and2x+y+z=4inthefirstoctantis
Answered by mr W last updated on 17/Aug/24
V=(1/6)(2×4×4−2×2×1)=((14)/3)
V=16(2×4×42×2×1)=143
Commented by universe last updated on 17/Aug/24
sir can u explain your method little bit  and how tw solve by integration method
sircanuexplainyourmethodlittlebitandhowtwsolvebyintegrationmethod
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)
theeqn.ofblueplaneisxa+yb+zc=1thevolumeundertheplaneisV1=ab2×c3=abc6similarlythevolumeundertheredplanexp+yq+zr=1isV2=pqr6thevolumebetweenblueplaneandredplaneisV=V1V2=16(abcpqr).inthisexampletheredplaneisx+y+2z=2,i.e.x2+y2+z=1andtheblueplaneis2x+y+z=4,i.e.x2+y4+z4=1.thereforeV=16(2×4×42×2×1)=143
Commented by mr W last updated on 18/Aug/24
Commented by mr W last updated on 18/Aug/24
there is no need to apply integration.
thereisnoneedtoapplyintegration.
Commented by universe last updated on 18/Aug/24
thank u sir
thankusir

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