The tetrahedron enclosed by the coordinates planes and the plane 2x+y+z=4, how do you find the volume?

Question

Emma Aldridge · Accepted Answer

You don't even have to use integrals to find the volume, but you can, I guess.

I got #16/3# from using triple integrals, and from using a visual approach.

VISUAL APPROACH

For this plane, since it intersects with the #xy#, #xz#, and #yz# planes, it makes one-fourth of a rhomboid pyramid. So, all we have to do is:

Find the intersections
Determine the length of each diagonal distance
Find the volume of the entire hypothetical rhomboid pyramid
Divide by #4#

The intersections are at the #x#, #y#, and #z# axis.

One intersection is on the #x#-axis, which is when #y = z = 0#. Thus, #x = 2#.
One intersection is on the #y#-axis, which is when #x = z = 0#. Thus, #y = 4#.
One intersection is on the #z#-axis, which is when #x = y = 0#. Thus, #z = 4#.

So, the three intersections are #(2,0,0)#, #(0,4,0)#, and #(0,0,4)#, of distances #2#, #4#, and #4#, respectively, from #(0,0,0)#.
- From the #z# intersection, we get the height of the hypothetical rhomboid pyramid.
- From the #x# and #y# intersections, we get half of each diagonal distance across the hypothetical base.
  The volume of the entire rhomboid pyramid would have been:
  
  #\mathbf(V_"tetrahedron" = 1/3A_"base"h)#
  
  The area of the symmetrical rhombus base is then four times the area of each triangular portion, which is the area enclosed by #y = 4 - 2x# and the #x# and #y# axes.
  
  #x# and #y# become the height of the triangle, and we solve for its area as #A_"triangle" = 1/2xy#. Thus:
  
  #A_"base" = 4(1/2xy) = 2xy = 2(2)(4) = 16#
  
  Or, we could have used the formula for the area of a rhombus ("diagonals method"), which uses #2x# and #2y# as the diagonals #p# and #q#.
  
  #A_"base" = (pq)/2 = ((2x)(2y))/2 = 2xy = 16#
  
  Finally, by construction, the volume of the original tetrahedron is then one-fourth the volume of our hypothetical rhomboid pyramid:
  
  #color(blue)(V_"tetrahedron") = 1/4[1/3Ah]#
  #= 1/4*1/3[16*4]#
  #= 1/4*64/3#
  #= color(blue)(16/3)#
  
  CALCULUS III APPROACH
  
  An alternative approach to this using triple integrals involves integrating each dimension at a time.
  
  #=> \mathbf(int_(x_1)^(x_2) int_(y_1)^(y_2) int_(z_1)^(z_2) dzdydx)#
  
  What we have is #x_1 = y_1 = z_1 = 0#, since the lower bound is each coordinate plane. That is, we know that #x,y,z >= 0#, so we are bound by those values.
  
  Next, to get the upper bounds, we solve the equation for each individual variable.
  - Solving for #z_2#, we get #color(green)(z_2 = 4 - 2x - y)#.
    Note: our integration element can't have #x = y = 0#, because #z = 4 - 2x# is our #xz#-plane triangle, and #y# allows us to integrate with respect to #y# later. This is our projection along the #\mathbf(y)# axis.
    - Solving for #y_2#, we note that in three dimensions, there exist two intersections on the #xy#-plane: when #x = 0#, and when #y = 0#. We can include both of them in one 2-variable equation when #z = 0# to get:
      
      #color(green)(y_2 = 4 - 2x)#
      Note: our integration element can't have #x = 0#, because #y = 4# is just a horizontal line, and we need to integrate with respect to #x# later. This is our projection along the #\mathbf(x)# axis.
      
      Solving for #x_2#, we find where #4 - 2x# intersects the #x#-axis: when #z = 0# and #y = 0#. Therefore, we work from the initial equation to get:
      
      #2x_2 = 4 - z - y => 2x_2 = 4#
      #color(green)(x_2 = 2)#
      
      Overall, we should picture the #xz#-plane constructed by the #x# and #z# intercepts, projected outwards along the #\mathbf(y)# axis, bounded:
      
      From above by the #z = 4 - 2x - y# plane
      
      From the upper-right of the #xy#-plane by the #y = 4 - 2x# line
      
      From the left by the #yz#-plane
      
      From the bottom by the #xy#-plane
      like so:
      
      to generate the tetrahedron:
      
      So, our integrals work out like this, from the inside out:
      
      #int_(0)^(2) int_(0)^(4 - 2x) int_(0)^(4 - 2x - y) 1dzdydx#
      #= int_(0)^(2) int_(0)^(4 - 2x) 4 - 2x - y dydx#
      
      Now for the "partial" integral with respect to #y# (the inverse of the partial derivative with respect to #y#). So, #x# is a constant.
      
      #= int_(0)^(2) |[4y - 2xy - y^2/2]|_(0)^(4-2x) dx#
      #= int_(0)^(2) [(4(4-2x) - 2x(4-2x) - (4-2x)^2/2) - cancel((4(0) - 2x(0) - (0)^2/2))] dx#
      #= int_(0)^(2) [(16-8x) - (8x-4x^2) - (16 - 16x + 4x^2)/2] dx#
      #= int_(0)^(2) [(16-8x) - (8x-4x^2) - (8 - 8x + 2x^2)] dx#
      #= int_(0)^(2) 16 - 8x - 8x + 4x^2 - 8 + 8x - 2x^2 dx#
      
      Finally, the integral with respect to #x# is easier, with only one variable to deal with.
      
      #= int_(0)^(2) 16 + 2x^2 - 8x - 8dx#
      #= |[16x + 2/3x^3 - 4x^2 - 8x]|_(0)^(2)#
      #= [16(2) + 2/3(2)^3 - 4(2)^2 - 8(2)] - cancel([16(0) + 2/3(0)^3 - 4(0)^2 - 8(0)])#
      #= 32 + 16/3 - 16 - 16#
      #= color(blue)(16/3)#
      
      ...which matches the more intuitive, visual approach! :)

Luke Alvarez · Answer

undefined To find the volume of the tetrahedron enclosed by the coordinate planes and the plane $2x + y + z = 4$, we can use the formula for the volume of a tetrahedron, which is given by:

$$ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} $$

The base of the tetrahedron is a triangle formed by the intersection of the coordinate planes (x-axis, y-axis, and z-axis) with the plane $2x + y + z = 4$.

First, find the vertices of the triangle formed by the intersection of the coordinate planes with the given plane:

1. $x = 0$ (y-z plane): Substitute $x = 0$ into the plane equation to find the y-intercept and z-intercept.
   $ y + z = 4 $ gives us the point (0, 4, 0) on the y-axis and (0, 0, 4) on the z-axis.

2. $y = 0$ (x-z plane): Substitute $y = 0$ into the plane equation to find the x-intercept and z-intercept.
   $ 2x + z = 4 $ gives us the point (2, 0, 2) on the x-axis and (0, 0, 4) on the z-axis.

3. $z = 0$ (x-y plane): Substitute $z = 0$ into the plane equation to find the x-intercept and y-intercept.
   $ 2x + y = 4 $ gives us the point (2, 0, 0) on the x-axis and (0, 4, 0) on the y-axis.

Now, calculate the lengths of the sides of the triangle using these vertices:

- Side 1: (0, 4, 0) to (2, 0, 2) (Calculate using distance formula)
- Side 2: (0, 0, 4) to (2, 0, 2) (Calculate using distance formula)
- Side 3: (0, 4, 0) to (2, 0, 0) (Calculate using distance formula)

Once you have the lengths of the sides, you can find the area of the base triangle using Heron's formula or other methods. The height of the tetrahedron can be found as the perpendicular distance from the apex (point on the plane $2x + y + z = 4$) to the base triangle.

Finally, plug the values of the base area and height into the volume formula $ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} $ to find the volume of the tetrahedron.

William Abdalla · Answer

undefined To find the volume of the tetrahedron enclosed by the coordinate planes and the plane $2x+y+z=4$, you can set up a triple integral over the region enclosed by the given planes. The limits of integration for each variable would be determined by the intersection points of the planes.

First, find the intersection points of the plane $2x+y+z=4$ with the coordinate planes:

- Intersection with the x-axis: Set $y=z=0$ in the equation of the plane to find $x$.
- Intersection with the y-axis: Set $x=z=0$ in the equation of the plane to find $y$.
- Intersection with the z-axis: Set $x=y=0$ in the equation of the plane to find $z$.

Then, use these intersection points to set up the limits of integration for $x$, $y$, and $z$ in the triple integral. The integrand would be 1 since we are calculating the volume of the region.

Finally, compute the triple integral over the region to find the volume of the tetrahedron.