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Question Number 182368 by universe last updated on 08/Dec/22
    find volume of region in  R^3   given by       3∣x∣ + 4∣y∣ +3∣z∣ ≤12  is
$$\:\:\:\:\mathrm{find}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{region}\:\mathrm{in}\:\:\mathbb{R}^{\mathrm{3}} \:\:\mathrm{given}\:\mathrm{by}\: \\ $$$$\:\:\:\:\mathrm{3}\mid\mathrm{x}\mid\:+\:\mathrm{4}\mid\mathrm{y}\mid\:+\mathrm{3}\mid\mathrm{z}\mid\:\leqslant\mathrm{12}\:\:\mathrm{is} \\ $$
Answered by mr W last updated on 08/Dec/22
((∣x∣)/4)+((∣y∣)/3)+((∣z∣)/4)≤1  V=8×((4×3×4)/6)=64
$$\frac{\mid{x}\mid}{\mathrm{4}}+\frac{\mid{y}\mid}{\mathrm{3}}+\frac{\mid{z}\mid}{\mathrm{4}}\leqslant\mathrm{1} \\ $$$${V}=\mathrm{8}×\frac{\mathrm{4}×\mathrm{3}×\mathrm{4}}{\mathrm{6}}=\mathrm{64} \\ $$
Commented by universe last updated on 09/Dec/22
sorry sir i dont get it  please explain little more
$${sorry}\:{sir}\:{i}\:{dont}\:{get}\:{it} \\ $$$${please}\:{explain}\:{little}\:{more} \\ $$
Commented by mr W last updated on 09/Dec/22
Commented by mr W last updated on 09/Dec/22
the part of the region in octant I  is the pyramid as shown. it′s volume  is ((4×3)/2)×(4/3)=((4×3×4)/6)  for the total region in all 8 octants the  volume is V=8×((4×3×4)/6)=64
$${the}\:{part}\:{of}\:{the}\:{region}\:{in}\:{octant}\:{I} \\ $$$${is}\:{the}\:{pyramid}\:{as}\:{shown}.\:{it}'{s}\:{volume} \\ $$$${is}\:\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{4}×\mathrm{3}×\mathrm{4}}{\mathrm{6}} \\ $$$${for}\:{the}\:{total}\:{region}\:{in}\:{all}\:\mathrm{8}\:{octants}\:{the} \\ $$$${volume}\:{is}\:{V}=\mathrm{8}×\frac{\mathrm{4}×\mathrm{3}×\mathrm{4}}{\mathrm{6}}=\mathrm{64} \\ $$
Commented by mr W last updated on 09/Dec/22
Commented by universe last updated on 09/Dec/22
thanks sir
$${thanks}\:{sir} \\ $$

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