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

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:

1. Find the intersections
2. Determine the length of each diagonal distance
3. Find the volume of the entire hypothetical rhomboid pyramid
4. 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! :)