Volume integral - finding the regionI need to find the volume of an object set with the following function: x + y + z = 1And all three axis.So I converted the function into z = − x − y + 1, and it gave me kind of a clepsydra, crossing x,y plane in x = y = 1.So the region seems to be the circle x 2 + y 2 = 1.However, changing x to r, and y to ϕ gives me wrong result π, and it should be 1/6.Is the region I've came up with correct?

Question

Volume integral - finding the regionI need to find the volume of an object set with the following function:  x  +  y  +  z  =  1And all three axis.So I converted the function into   z  =  −  x  −  y  +  1, and it gave me kind of a clepsydra, crossing x,y plane in   x  =  y  =  1.So the region seems to be the circle       x          2        +      y          2        =  1.However, changing x to r, and y to   ϕ gives me wrong result   π, and it should be 1/6.Is the region I&#039;ve came up with correct?

prelatiuvq · Accepted Answer

Explanation:                              V                                    ≡                                            ∭                                                                    (                                                              0                      ,                      1                                                            )                                                                            3                                                                                                  [                                            x                +                y                +                z                &amp;lt;                1                                          ]                                      d                        x                                      d                        y                                      d                        z                                              =                                       ∫                      0                                1                                    ∫                      0                                1                                    ∫                      0                                1                                    [                                x            &amp;lt;            1            −            y            −            z                              ]                          d                x                          d                y                          d                z                            =                                       ∫                      0                                1                                    ∫                      0                                1                                    [                                0            &amp;lt;            1            −            y            −            z            &amp;lt;            1                              ]                          ∫                      0                                1            −            y            −            z                                    d                x                  d                y                          d                z                            =                                       ∫                      0                                1                                    ∫                      0                                1                                    [                                −            z            &amp;lt;            y            &amp;lt;            1            −            z                              ]                          (                                1            −            y            −            z                              )                          d                y                            =                                       ∫                      0                                1                                    ∫                      0                                1                                    [                                y            &amp;lt;            1            −            z                              ]                          (                                1            −            y            −            z                              )                          d                y                          d                z                            =                                       ∫                      0                                1                                    ∫                      0                                1            −            z                                    (                                1            −            y            −            z                              )                          d                y                  d                z                            =                                                   1            2                                    ∫                      0                                1                                                (                                      1              −              z                                    )                                2                                    d                z        =                                                                                      1                  6

logosdepmpe · Accepted Answer

Step 1First of all, the region of interest here is a right tetrahedron, with a right triangle base of side length 1 and height 1, so the volume is       1    3    ⋅      1    2    (  1  )  (  1  )  =      1    6  .The volume integral may be done by integrating over, say, x first, then y, then z:  V  =      ∫    0    1    d  z        ∫    0          1      −      z        d  y        ∫    0          1      −      y      −      z        d  xStep 2Doing the innermost integral:  V  =      ∫    0    1    d  z        ∫    0          1      −      z        d  y    (  1  −  y  −  z  )  =      ∫    0    1    d  z        [    (    1    −    z          )      2        −          1      2        (    1    −    z          )      2        ]    =      1    2        ∫    0    1    d  z    (  1  −  z      )    2  Doing the final integral,   V  =      1    2    ⋅      1    3    ⋅  1  =      1    6

Find the volume of an object set with the following function: x+y+z=1.

Answered question

Answer & Explanation

New Questions in High school geometry