Triple Integral Calculation Of F(x, Y, Z) = X² Over A Tetrahedron

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Hey guys! Today, we're diving into the exciting world of triple integrals and how to find them for functions over specific regions in 3D space. Specifically, we're going to tackle a problem where we need to find the integral of the function f(x, y, z) = x² over the region S, which is a tetrahedron bounded by the plane 12x + 20y + 15z = 60 and the coordinate planes. Sounds like a mouthful, right? But don't worry, we'll break it down step by step!

Understanding the Problem

Before we jump into the calculations, let's make sure we understand what we're dealing with. The keyword here is triple integral. Think of it as the 3D version of a regular integral, which calculates the area under a curve. A triple integral, on the other hand, calculates the "hypervolume" under a surface in 3D space. In our case, the "surface" is defined by the function f(x, y, z) = x², and we want to find the hypervolume over the region S.

The region S is a tetrahedron, which is a pyramid with a triangular base. It's formed by the intersection of the plane 12x + 20y + 15z = 60 and the coordinate planes (the xy, xz, and yz planes). Imagine slicing off a corner of a room – that's essentially what our tetrahedron looks like. Visualizing this region is super important for setting up the integral correctly.

So, our mission is clear: find the triple integral of f(x, y, z) = x² over this tetrahedron S. Let's get to it!

Setting Up the Triple Integral

The first crucial step in solving this problem is setting up the triple integral. This involves determining the limits of integration for x, y, and z. To do this effectively, we need to understand how the region S is bounded in each direction.

  1. Finding the Intercepts:

    Let's start by finding the intercepts of the plane 12x + 20y + 15z = 60 with the coordinate axes. This will give us the vertices of our tetrahedron.

    • When y = 0 and z = 0, we have 12x = 60, so x = 5. This gives us the point (5, 0, 0).
    • When x = 0 and z = 0, we have 20y = 60, so y = 3. This gives us the point (0, 3, 0).
    • When x = 0 and y = 0, we have 15z = 60, so z = 4. This gives us the point (0, 0, 4).

    These three points, along with the origin (0, 0, 0), are the vertices of our tetrahedron. These vertices are incredibly valuable for defining the boundaries of our region.

  2. Determining the Limits of Integration:

    Now comes the slightly tricky part: figuring out the limits of integration. There are different orders of integration we can choose (dx dy dz, dz dy dx, etc.), and the best choice often depends on the specific problem. For this tetrahedron, let's try integrating with respect to z first, then y, and finally x (dz dy dx).

    • Limits for z:

      For a given (x, y), z varies from the xy-plane (z = 0) up to the plane 12x + 20y + 15z = 60. Solving the plane equation for z, we get:

      z = (60 - 12x - 20y) / 15 = 4 - (4/5)x - (4/3)y

      So, the limits for z are 0 ≤ z ≤ 4 - (4/5)x - (4/3)y.

    • Limits for y:

      Next, we need to find the limits for y. Imagine projecting the tetrahedron onto the xy-plane. The projection is a triangle bounded by the x-axis, the y-axis, and the line formed by setting z = 0 in the plane equation:

      12x + 20y = 60

      Solving for y, we get:

      y = (60 - 12x) / 20 = 3 - (3/5)x

      So, for a given x, y varies from 0 to 3 - (3/5)x. The limits for y are 0 ≤ y ≤ 3 - (3/5)x.

    • Limits for x:

      Finally, x varies from 0 to its intercept with the x-axis, which we found earlier to be x = 5. So, the limits for x are 0 ≤ x ≤ 5.

  3. Writing the Triple Integral:

    Now we can write the triple integral:

    ∫₀⁵ ∫₀⁽³⁻⁽³/₅⁾ˣ⁾ ∫₀⁽⁴⁻⁽⁴/₅⁾ˣ⁻⁽⁴/₃⁾ʸ⁾ x² dz dy dx

    This integral represents the hypervolume we're trying to find. It looks intimidating, but we'll tackle it one integral at a time.

Evaluating the Triple Integral

Alright, guys, let's get our hands dirty and evaluate the triple integral we just set up. Remember, we'll be working from the inside out, just like peeling an onion (a mathematical onion, of course!).

  1. Integrating with Respect to z:

    The innermost integral is with respect to z:

    ∫₀⁽⁴⁻⁽⁴/₅⁾ˣ⁻⁽⁴/₃⁾ʸ⁾ x² dz

    Since x² is constant with respect to z, this is a simple integral:

    x² ∫₀⁽⁴⁻⁽⁴/₅⁾ˣ⁻⁽⁴/₃⁾ʸ⁾ dz = x² [z]₀⁽⁴⁻⁽⁴/₅⁾ˣ⁻⁽⁴/₃⁾ʸ⁾ = x² (4 - (4/5)x - (4/3)y)

    So, after the first integration, our triple integral becomes a double integral:

    ∫₀⁵ ∫₀⁽³⁻⁽³/₅⁾ˣ⁾ x² (4 - (4/5)x - (4/3)y) dy dx

  2. Integrating with Respect to y:

    Now we integrate with respect to y:

    ∫₀⁽³⁻⁽³/₅⁾ˣ⁾ x² (4 - (4/5)x - (4/3)y) dy

    Distribute the x² and integrate term by term:

    ∫₀⁽³⁻⁽³/₅⁾ˣ⁾ (4x² - (4/5)x³ - (4/3)x²y) dy = [4x²y - (4/5)x³y - (2/3)x²y²]₀⁽³⁻⁽³/₅⁾ˣ⁾

    Now, plug in the limits of integration for y:

    [4x²(3 - (3/5)x) - (4/5)x³(3 - (3/5)x) - (2/3)x²(3 - (3/5)x)²] - 0

    This looks messy, but we can simplify it. After some algebraic gymnastics (which I'll spare you the details of, but feel free to work it out yourself!), we get:

    (6/25)x⁴ - (6/5)x³ + 12x²

    So, our double integral becomes a single integral:

    ∫₀⁵ [(6/25)x⁴ - (6/5)x³ + 12x²] dx

  3. Integrating with Respect to x:

    Finally, we integrate with respect to x:

    ∫₀⁵ [(6/25)x⁴ - (6/5)x³ + 12x²] dx = [(6/125)x⁵ - (3/10)x⁴ + 4x³]₀⁵

    Plug in the limits of integration for x:

    [(6/125)(5)⁵ - (3/10)(5)⁴ + 4(5)³] - 0 = 375 - 187.5 + 500 = 180

    Therefore, the final result of the triple integral is 180.

The Answer and Its Meaning

So, after all that work, we've found that the triple integral of f(x, y, z) = x² over the tetrahedron S is 180. But what does this number actually mean?

Remember, we said that the triple integral represents a "hypervolume." In this case, it's the hypervolume under the surface f(x, y, z) = x² and above the region S. It's a bit hard to visualize, but think of it as a four-dimensional volume. This numerical value, 180, quantifies this hypervolume.

In simpler terms, we've calculated the "weighted volume" of the tetrahedron, where the weight at each point is given by the function x². This kind of calculation has applications in various fields, such as physics (calculating mass or moment of inertia) and engineering (analyzing stress distributions).

Key Takeaways

  • Triple integrals are used to calculate hypervolumes in 3D space.
  • Setting up the integral correctly is crucial, and it involves determining the limits of integration based on the region's boundaries.
  • Evaluating the integral is done step-by-step, integrating from the inside out.
  • The result of a triple integral represents a hypervolume or a weighted volume, depending on the context.

Conclusion

So, guys, we've successfully navigated the world of triple integrals and found the integral of f(x, y, z) = x² over a tetrahedron. It was a challenging journey, but we broke it down into manageable steps. Remember, the key to mastering triple integrals is practice, practice, practice! Keep working on problems, and you'll become a triple integral pro in no time.

Now go forth and conquer those integrals!