Evaluate the triple integral
∫∫∫ExyzdV
where E is the solid: 0≤z≤6, 0≤y≤z, 0≤x≤y
.
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.
To evaluate the triple integral ∫∫∫ExyzdV over the given solid E, we need to set up the integral with the appropriate limits of integration.
The solid E is defined with the following bounds:
0 ≤ z ≤ 6
0 ≤ y ≤ z
0 ≤ x ≤ y
Therefore, let's evaluate the integral using the given bounds.
We start by integrating with respect to x first, then y, and finally z.
∫∫∫ExyzdV = ∫[0,6]∫[0,z]∫[0,y]Exyzdxdydz
First, we integrate with respect to x:
∫[0,y]Exyzdx = [xy^3/2]/2 = (xy^3/2)/2
Next, we integrate the previous result with respect to y:
∫[0,z] (xy^3/2)/2 dy = ∫ [(xz^3/2)/2 - 0] dy = (xz^3/2)/2
Finally, we integrate the previous result with respect to z:
∫[0,6] (xz^3/2)/2 dz = [(2/5)xz^5/2]/2 = (1/5)xz^5/2
Therefore, the triple integral ∫∫∫ExyzdV is given by the expression:
∫[0,6]∫[0,z]∫[0,y]Exyzdxdydz = (1/5)xz^5/2 evaluated for x = 0 to y, y = 0 to z, and z = 0 to 6.
To find the exact value of this integral, we substitute the limits of integration into the expression:
[(1/5)(y)(6)^(5/2)] evaluated for y = 0 to z
= (1/5)(z)(6)^(5/2) - (1/5)(0)(6)^(5/2)
= (1/5)(z)(6)^(5/2)
Finally, we substitute the upper limit 6 into the expression:
= (1/5)(6)(6)^(5/2)
= (6/5)(6)^(5/2)
= (6/5)(36√6)
= 216√6/5
Therefore, the value of the triple integral ∫∫∫ExyzdV over the solid E is 216√6/5.