Evaluate the triple integral _E (xy)dV where E is a solid tetrahedron with vertices (0,0,0), (4,0,0), (0,1,0), (0,0,4)
To evaluate the triple integral ∫∫∫_E (xy)dV, where E is a solid tetrahedron with vertices (0,0,0), (4,0,0), (0,1,0), (0,0,4), we will use a method called the setup and change of variables technique.
Step 1: Setting up the Integral Bounds
In order to integrate over the solid tetrahedron, we need to determine the limits of integration for each variable (x, y, z). Let's go through each variable one-by-one.
For x:
Since the tetrahedron is bound by the x-axis and the plane y = 1 - (x/4), the limits for x will be from 0 to 4.
For y:
Since the tetrahedron is bound by the y-axis, the plane x = 0, and the plane x + 4y + z = 4, we need to find the limits for y when z is fixed.
- If z = 0, the plane equation becomes x + 4y = 4. Solving for y, we get y = (4 - x)/4.
- If z = 4 - x, the plane equation becomes x + 4y + (4 - x) = 4, which simplifies to y = 0.
Therefore, the limits for y will be from 0 to (4 - x)/4.
For z:
Since the tetrahedron is bound by the xy-plane and the plane x + 4y + z = 4, the limits for z will be from 0 to 4 - x - 4y.
Step 2: Changing Variables
To simplify the integral, we will introduce a change of variables. Let's define new variables u, v, and w, such that:
- x = u
- y = v/4
- z = w
Now, let's apply the change of variables to the integral.
Step 3: Rewriting the Integral
The integral becomes: ∫∫∫_E (xy)dV = ∫∫∫_E (u)(v/4)dudv(dw/dx)dv
Step 4: Determining the New Limits
We need to express the limits of integration in terms of u, v, and w. Using the same reasoning as before, the new limits become:
- For u: from 0 to 4
- For v: from 0 to 4 - u
- For w: from 0 to 4 - u - v
Step 5: Changing the Integrals
Now that we have our new integral bounds and variables, we can rewrite the triple integral as follows:
∫∫∫_E (xy)dV = ∫∫∫_E (u)(v/4)dudv(dw/dx)dv
= ∫₀⁴ ∫₀^(4-u) ∫₀^(4-u-v) (u)(v/4)(1) dwdvdudx
Step 6: Evaluating the Integral
To evaluate the triple integral, we integrate with respect to w, then v, and finally u.
∫₀⁴ ∫₀^(4-u) ∫₀^(4-u-v) (u)(v/4)(1) dwdvdudx
= ∫₀⁴ ∫₀^(4-u) [(1/2)v²(4-u-v)]₀^(4-u-v) dv dudx
= ∫₀⁴ ∫₀^(4-u-y) [(1/2)v²(4-u-v)] dv dudx
= ∫₀⁴ [(1/6)v³(4-u-v)]₀^(4-u) dudx
= ∫₀⁴ [(64-18u+u²-2u³+u⁴)/30] dudx
= (1/30) ∫₀⁴ (u⁴-2u³+u²-18u+64) dudx
= (1/30) [(1/5)u⁵-(1/2)u⁴+(1/3)u³-(9/2)u²+64u]₀⁴
= (1/30) [(1/5)(4⁵)-(1/2)(4⁴)+(1/3)(4³)-(9/2)(4²)+64(4)]
Evaluating this expression will give you the final numerical value of the triple integral.