Evaluate the triple integral

Question

∫∫∫ExyzdV
where E is the solid: 0≤z≤6, 0≤y≤z, 0≤x≤y
.

Answer 1

We will evaluate the triple integral ∫∫∫Exyz dV by considering the bounds of integration for each variable.

The solid E is defined by the following inequalities:
0 ≤ z ≤ 6
0 ≤ y ≤ z
0 ≤ x ≤ y

Let's start by integrating with respect to x. Since x varies from 0 to y, the limits of integration for x with respect to y are 0 to y.

Next, we integrate with respect to y. Since y varies from 0 to z, the limits of integration for y with respect to z are 0 to z.

Finally, we integrate with respect to z. Since z varies from 0 to 6, the limits of integration for z are 0 to 6.

Therefore, the triple integral becomes:

∫(0 to 6) ∫(0 to z) ∫(0 to y) Exyz dx dy dz

To evaluate this integral, we will calculate each integration step by step.

First, let's integrate with respect to x:
∫(0 to y) Exyz dx = E * ∫(0 to y) (xyz) dx

Integrating (xyz) with respect to x gives (1/2)xyz^2.

Therefore, the integral becomes:
E * ∫(0 to y) (1/2)xyz^2 dx = E * (1/2)xy^2z^2

Next, let's integrate with respect to y:
∫(0 to z) E * (1/2)xy^2z^2 dy = E * (1/2) x * [y^3/3] (0 to z) = E * (1/6) x * z^3

Finally, let's integrate with respect to z:
∫(0 to 6) E * (1/6) x * z^3 dz = E * (1/24) x * [z^4/4] (0 to 6) = E * (1/96) x * 6^4

Simplifying, we have:
E * (1/96) x * 6^4 = E * (1/96) x * 1296

And, since we have integrated with respect to x, we can substitute the bounds of integration:
E * (1/96) * [0 to y] * 1296 = E * (1/96) * y * 1296

Now, we substitute the bounds of integration for y:
E * (1/96) * 6 * 1296 - E * (1/96) * 0 * 1296

Simplifying further, we have:
E * (1/96) * 6 * 1296 = E * 9 * 18 = 162E

Therefore, the triple integral ∫∫∫Exyz dV evaluates to 162E.