Finding volume of a shape using double integralI'm trying to find the volume of a given shape: V = { x + y + z ≤ 1 x ≥ 0 , y ≥ 0 , z ≥ 0 using double integral. Unfortunately I don't know how to start, namely: z = ( 1 − y − x ) 2 and now what should I do? Wolfram can't even plot this function, I'm unable to imagine how it looks like...Would it be simpler with a triple integral?

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Finding volume of a shape using double integralI&#039;m trying to find the volume of a given shape:  V  =      {                                        x                    +                      y                    +                      z                    ≤          1                                      x          ≥          0          ,                     y          ≥          0          ,                     z          ≥          0                        using double integral. Unfortunately I don&#039;t know how to start, namely:  z  =  (  1  −      y    −      x        )          2      and now what should I do? Wolfram can&#039;t even plot this function, I&#039;m unable to imagine how it looks like...Would it be simpler with a triple integral?

Skylar Beard · Accepted Answer

Step 1By using curve expert program this is an approximation formula in the first quadrant:  v  =  A  B  C  (  a  b  +  c      n    d    )      /    (  b  +      n    d    )for   n  =  {  .3  ,  .  .  .  .  ,  1  }  a  =  −  7.996  ×      10          −      4        b  =  9.2  ×      10          −      1        c  =  3.198  ×      10          −      1        d  =  4.7Step 2for   n  =  {  1  ,  .  .  .  .  ,  100  }  a  =  −  1.77  ×      10          −      1        b  =  2.43  c  =  1  d  =  1.83so for   n  =  .5, volume approximated   =  0.012.

yongenelowk · Accepted Answer

Step 1The volume can be calculated as       ∫    0    1        ∫    0          (      1      −              x                    )        2              (  1  −      x    −      y        )    2    d  y  d  x  ..Maple was able to find the inner integral (function of x), but unable to finish the evaluation by integrating that from 0 to 1. However a numeric approximation gave 0.011111111... which looks like it is 1/90.Step 2EDIT: If the substitutions   x  =      r    2    ,  y  =      s    2    ,  z  =      t    2   are made, noting that   d  x  =  2  r     d  r,   d  y  =  2  s     d  s, and   d  z  =  2  t     d  t, the volume element in the r,s,t system is 8rst dt ds dr, and the volume becomes      ∫    0    1        ∫    0          1      −      r            ∫    0          1      −      r      −      s        8  r  s  t     d  t     d  s     d  r  =      1    90    ,where the integration in the r,s,t system is not complicated by squareroots.Christian Blatter did exactly this substitution in his answer, and I just noticed that after doing the above edit.

Find the volume of a given shape: V={(\sqrt{x}+,\sqrt{y}+,\sqrt{z}\leq1),(x \geq 0, y\geq0,z\geq0):}

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