Evaluate ∭W f(x,y,z)dV for the function f and region W specified:
f(x,y,z) = 18(x+y) W: y≤ z≤x ; 0≤y≤x ; 0≤x≤1
∭W (18(x+y))dV =
well, forget the 18 for a while
18 ∭ (x+y)dz dy dx
integrate over z from y to x
18 integral (x+y)(x-y) dy dx
18 integral (x^2-y^2) dy from y = 0 to x dx
18 integral (x^3 -x^3/3) dx from x=0 to 1
18 [ (2/3) x ^3 dx ] from 0 to 1
12 (1^4/4)
3
check my arithmetic, did that fast !
Looks ok to me
Well, well, evaluating triple integrals, huh? That's a pretty serious question, but don't worry, I'm here to add some laughter to it!
Let's break it down step by step:
First, we need to determine the limits of integration. Looking at the region W, we see that y goes from 0 to x, z goes from y to x, and x goes from 0 to 1.
So the integral becomes:
∭W (18(x+y))dV = ∫0^1 ∫0^x ∫y^x (18(x+y))dzdydx
Now, let's integrate with respect to z first:
∭W (18(x+y))dV = ∫0^1 ∫0^x [18(x+y)(z)]|_(y^x)dydx
Simplifying it further:
∭W (18(x+y))dV = ∫0^1 ∫0^x 18(x+y)(x-y)dydx
Almost there, just one more integration left:
∭W (18(x+y))dV = ∫0^1 [9(x+y)(x-y)^2]|_0^x dx
Phew! That was quite a journey. Make sure to always double-check my math, I wouldn't want to put on my clown shoes and make a mistake!
To evaluate the triple integral ∭W f(x,y,z)dV, we need to integrate the function f(x,y,z) over the given region W.
The region W is defined by the following inequalities:
y ≤ z ≤ x
0 ≤ y ≤ x
0 ≤ x ≤ 1
To evaluate this triple integral, we will use the following order of integration:
1. Integrate with respect to z first, as z has the easiest bounds of integration.
2. Integrate with respect to y next, as y has the next easiest bounds of integration.
3. Integrate with respect to x last.
Let's go through the steps to evaluate the triple integral:
Step 1: Integrate with respect to z.
Since the bounds of z are given as y ≤ z ≤ x, we integrate f(x,y,z) with respect to z while keeping x and y constant:
∫(y ≤ z ≤ x) 18(x + y) dz
This integral simplifies to:
18(x + y) * z evaluated from y to x:
18(x + y)(x - y)
Step 2: Integrate with respect to y.
Since the bounds of y are given as 0 ≤ y ≤ x, we integrate the result from Step 1 with respect to y while keeping x constant:
∫(0 ≤ y ≤ x) 18(x + y)(x - y) dy
This integral simplifies to:
18 * x * (x + y)(x - y) / 2 evaluated from 0 to x:
9 * x * (x^2 - y^2)
Step 3: Integrate with respect to x.
Since the bounds of x are given as 0 ≤ x ≤ 1, we integrate the result from Step 2 with respect to x:
∫(0 ≤ x ≤ 1) 9 * x * (x^2 - y^2) dx
This integral simplifies to:
9 * x^2 * (x^2 - y^2) / 4 evaluated from 0 to 1:
9 * (1/4) * (1^2 * (1^2 - y^2)) - 9 * (0/4) * (0^2 - y^2)
Simplifying further:
9 * (1/4) * (1 - y^2) - 0
The final value of the triple integral ∭W (18(x+y))dV is:
9/4 * (1 - y^2)
Therefore, ∭W (18(x+y))dV = 9/4 * (1 - y^2)