Question

Find the volume integral of \(x^{2} y\) over the tetrahedral volume bounded by the planes \(x=0, y=0, z=0\), and \(x+y+z=1\)

Short answer

The volume integral of \(x^{2} y\) over the given tetrahedral volume is evaluated using the bounds provided by the intersecting planes.

Step-by-step solution

Identify the Bounds

The tetrahedral volume is bounded by the planes: 1. \(x = 0\)2. \(y = 0\)3. \(z = 0\)4. \(x + y + z = 1\). These planes provide the limits for the integrations.

Set Up the Triple Integral

The volume integral of \(x^{2} y\) over the tetrahedron can be represented as:\[ \iiint\{V\} x^{2} y \, dV \qquad \text{ where } dV = dx \, dy \, dz.\]

Order of Integration

Choose the order of integration. Here, we will integrate with respect to \(z\) first, then \(y\), and \(x\) last:\[ \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} x^{2} y \, dz \, dy \, dx \]

Integrate with Respect to z

Integrate the inner integral with respect to \(z\):\[ \int_{0}^{1} \int_{0}^{1-x} \, \left[ x^{2} y z \right]_{z=0}^{z=1-x-y} \, dy \, dx \]This evaluates to:\[ \int_{0}^{1} \int_{0}^{1-x} x^{2} y (1 - x - y) \, dy \, dx \]

Simplify and Integrate with Respect to y

Distribute and integrate with respect to \(y\):\[ \int_{0}^{1} \int_{0}^{1-x} (x^{2} y - x^{3} y - x^{2} y^{2}) \, dy \, dx \]Break this into three separate integrals:\[ \int_{0}^{1} \int_{0}^{1-x} x^{2} y \, dy \, dx - \int_{0}^{1} \int_{0}^{1-x} x^{3} y \, dy \, dx - \int_{0}^{1} \int_{0}^{1-x} x^{2} y^{2} \, dy \, dx \]

Evaluate Each Integral

Evaluate each integral separately:1. For \(x^{2} y\): \[ \int_{0}^{1} \left[ \frac{x^{2} y^{2}}{2} \right]_{0}^{1-x} \, dx = \int_{0}^{1} \frac{x^{2} (1-x)^{2}}{2} \, dx \]2. For \(x^{3} y\): \[ \int_{0}^{1} \left[ \frac{x^{3} y^{2}}{2} \right]_{0}^{1-x} \, dx = \int_{0}^{1} \frac{x^{3} (1-x)^{2}}{2} \, dx \]3. For \(x^{2} y^{2}\): \[ \int_{0}^{1} \left[ \frac{x^{2} y^{3}}{3} \right]_{0}^{1-x} \, dx = \int_{0}^{1} \frac{x^{2} (1-x)^{3}}{3} \, dx \] Evaluate each of these integrals.

Sum the Results

Summing the results, combine the evaluated integrals to get the final answer for the volume integral.

Key concepts

Triple Integral

In multivariable calculus, the triple integral allows us to integrate a function over a three-dimensional region.
Specifically, it’s used to calculate volumes, among other applications.

By integrating a function three times, using respect to three different variables, we can capture the cumulative sum of the function’s values throughout the 3D space.

For example, to find the integral of the function \(f(x,y,z)\) over a 3D region \(V\), we write this as:
\[ \tripleint_{V} f(x,y,z) \, dV \]

This means:

• First, you integrate with respect to one variable (let's say \(z\)).
• Next, you take the result and integrate it with respect to another variable (let's say \(y\)).
• Finally, you integrate with respect to the last variable (let's say \(x\)).

Integration Bounds

When setting up a triple integral, defining the bounds of integration is critical.

These bounds describe the region over which you integrate, and this can vary based on the geometric boundaries defined in the problem.

For our exercise, the region is a tetrahedron bounded by the planes:

• \(x=0\)
• \(y=0\)
• \(z=0\)
• \(x+y+z=1\)

Using these planes, we form the limits for each variable, ensuring our integrals only cover the region inside the tetrahedron.

The order of integration determines the lower and upper bounds for each variable:

• First, we integrate with respect to \(z\) from \(0\) to \(1-x-y\).
• Then, we integrate with respect to \(y\) from \(0\) to \(1-x\).
• Finally, we integrate with respect to \(x\) from \(0\) to \(1\).

Putting it mathematically, the integral is expressed as:
\[ \tripleint_{V} f(x,y,z) \, dV = \tripleint \tripleint \tripleint{}_{0}^{1}{}_{0}^{1-x}{}_{0}^{1-x-y} f(x,y,z) \, dz \, dy \, dx \].

Tetrahedron

A tetrahedron is a polyhedron with four triangular faces, six straight edges, and four vertex corners.

It’s the simplest type of three-dimensional figure with flat faces.

The tetrahedron in our problem is bounded by the planes \(x=0\), \(y=0\), \(z=0\), and \(x+y+z=1\).

This specific tetrahedron lies within the first octant of the coordinate system, with one vertex at the origin (0,0,0) and the opposite vertex at (1,1,1).

The tetrahedral volume helps us define the region for our triple integral.

Understanding the shape and boundaries of this tetrahedron is crucial in determining the correct integration limits and ensuring the function encompasses the entire volume to get an accurate result.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions of multiple variables.

Instead of handling just one variable, it deals with the relationships between multiple variables.

Common concepts include partial derivatives, multiple integrals (like double and triple integrals), and gradients.

In multivariable calculus, we often explore how changes in one variable affect another within higher-dimensional spaces.

By utilizing techniques such as triple integrals, we can solve complex problems involving volume calculations and other three-dimensional aspects.

The concepts are fundamental in fields such as engineering, physics, computer graphics, and economics, where phenomena often depend on multiple variables interacting simultaneously.