Calculating Volume Of Solid Bounded By Planes X=0, Y=0, Z=0, And X+y+z=1

In the realm of three-dimensional geometry, determining the volume of a solid is a fundamental concept with wide-ranging applications. From engineering design to physics calculations, the ability to quantify the space occupied by a three-dimensional object is essential. This article delves into the process of calculating the volume of a specific solid bounded by the planes $x=0$, $y=0$, $z=0$, and $x+y+z=1$. This solid, a tetrahedron, presents an excellent case study for understanding how to apply integral calculus to solve geometric problems.

Our journey begins with a clear definition of the solid in question. The planes $x=0$, $y=0$, and $z=0$ represent the coordinate planes, effectively forming the three sides of a corner in three-dimensional space. The plane $x+y+z=1$ introduces the fourth side, creating a closed, three-dimensional figure. This figure is a tetrahedron, a polyhedron with four triangular faces, six edges, and four vertices. Visualizing this solid is the first crucial step in understanding how to approach the volume calculation. We can imagine this tetrahedron sitting neatly in the first octant of the Cartesian coordinate system, with its vertices at the origin (0,0,0) and the points (1,0,0), (0,1,0), and (0,0,1) on the coordinate axes. This mental picture will guide us as we set up the integral that will yield the volume. — North Las Vegas Utilities: Setup, Billing & Saving Tips

The method we will employ to calculate the volume is triple integration. This powerful tool from calculus allows us to sum up infinitesimally small volumes over the region of interest. The key to successful triple integration lies in correctly setting up the limits of integration. These limits define the boundaries of the solid in three-dimensional space and ensure that we account for every point within the tetrahedron. We will systematically determine these limits by considering how the variables $x$, $y$, and $z$ relate to each other within the given planes. The equation $x+y+z=1$ will play a central role in defining these relationships. By expressing $z$ in terms of $x$ and $y$, we can establish the upper bound for $z$ at any given point in the $xy$-plane. Similarly, by analyzing the projections of the tetrahedron onto the coordinate planes, we can determine the limits for $x$ and $y$. This careful setup is the foundation upon which the entire volume calculation rests.

The first step in calculating the volume of the tetrahedron involves setting up the triple integral. This requires a clear understanding of the region of integration, which is the solid bounded by the planes $x=0$, $y=0$, $z=0$, and $x+y+z=1$. The key to setting up the triple integral correctly is to determine the appropriate limits of integration for each variable ($x$, $y$, and $z$). These limits define the boundaries of the solid and ensure that we integrate over the entire region.

To determine the limits of integration, it's helpful to visualize the tetrahedron in the first octant. The plane $x+y+z=1$ intersects the coordinate axes at the points (1,0,0), (0,1,0), and (0,0,1). These points, along with the origin (0,0,0), form the vertices of the tetrahedron. We can use this information to define the range of each variable. Let's start with $z$. From the equation of the plane $x+y+z=1$, we can express $z$ as $z = 1 - x - y$. Since the solid is bounded by the plane $z=0$ and the plane $x+y+z=1$, the limits of integration for $z$ are from $0$ to $1-x-y$. This means that for any given values of $x$ and $y$, $z$ varies from the $xy$-plane ($z=0$) up to the plane $x+y+z=1$.

Next, we need to determine the limits of integration for $y$. To do this, we project the tetrahedron onto the $xy$-plane. This projection is a triangle bounded by the lines $x=0$, $y=0$, and $x+y=1$. The line $x+y=1$ is the intersection of the plane $x+y+z=1$ with the $xy$-plane ($z=0$). From this equation, we can express $y$ as $y = 1 - x$. Since the solid is bounded by the plane $y=0$ and the line $x+y=1$, the limits of integration for $y$ are from $0$ to $1-x$. This means that for any given value of $x$, $y$ varies from the $x$-axis ($y=0$) up to the line $x+y=1$.

Finally, we determine the limits of integration for $x$. The projection of the tetrahedron onto the $x$-axis is the interval from $0$ to $1$. This is because the plane $x+y+z=1$ intersects the $x$-axis at the point (1,0,0). Therefore, the limits of integration for $x$ are from $0$ to $1$. With the limits of integration for $x$, $y$, and $z$ determined, we can now set up the triple integral. The volume $V$ of the tetrahedron is given by the triple integral:

$$V = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} dz\, dy\, dx$$

This integral represents the summation of infinitesimally small volumes ($dz\, dy\, dx$) over the region defined by the limits of integration. The order of integration is important, as it reflects the way we have defined the limits for each variable. We first integrate with respect to $z$, then with respect to $y$, and finally with respect to $x$. This order corresponds to the way we have determined the limits, starting with $z$ as a function of $x$ and $y$, then $y$ as a function of $x$, and finally $x$ as a constant range.

Now that we have successfully set up the triple integral, the next step is to evaluate it. This involves performing the integration in the correct order, starting with the innermost integral and working our way outwards. The triple integral we need to evaluate is:

$$V = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} dz\, dy\, dx$$

Let's begin by evaluating the innermost integral, which is with respect to $z$:

$$\int_{0}^{1-x-y} dz = [z]_{0}^{1-x-y} = (1-x-y) - 0 = 1-x-y$$

So, after integrating with respect to $z$, our triple integral becomes a double integral:

$$V = \int_{0}^{1} \int_{0}^{1-x} (1-x-y) dy\, dx$$

Now, we evaluate the integral with respect to $y$. This means treating $x$ as a constant and integrating the expression $(1-x-y)$ with respect to $y$:

$$\int_{0}^{1-x} (1-x-y) dy = [(1-x)y - \frac{1}{2}y^2]_{0}^{1-x}$$

Substituting the limits of integration for $y$, we get:

$$[(1-x)(1-x) - \frac{1}{2}(1-x)^2] - [0] = (1-x)^2 - \frac{1}{2}(1-x)^2 = \frac{1}{2}(1-x)^2$$

So, after integrating with respect to $y$, our double integral becomes a single integral:

$$V = \int_{0}^{1} \frac{1}{2}(1-x)^2 dx$$

Finally, we evaluate the integral with respect to $x$. This involves integrating the expression $\frac{1}{2}(1-x)^2$ with respect to $x$:

$$\int_{0}^{1} \frac{1}{2}(1-x)^2 dx$$

To solve this, we can use a simple substitution. Let $u = 1-x$, so $du = -dx$. When $x = 0$, $u = 1$, and when $x = 1$, $u = 0$. The integral then becomes: — Pumas Vs Atlanta United: Epic Soccer Showdown

$$-\frac{1}{2} \int_{1}^{0} u^2 du = \frac{1}{2} \int_{0}^{1} u^2 du$$

Now, we can easily integrate $u^2$ with respect to $u$:

$$\frac{1}{2} [\frac{1}{3}u^3]_{0}^{1} = \frac{1}{2} (\frac{1}{3}(1)^3 - \frac{1}{3}(0)^3) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$$

Therefore, the volume $V$ of the tetrahedron is:

$$V = \frac{1}{6}$$

In conclusion, we have successfully calculated the volume of the solid bounded by the planes $x=0$, $y=0$, $z=0$, and $x+y+z=1$. By setting up and evaluating the triple integral, we found that the volume of this tetrahedron is $\frac{1}{6}$ cubic units. This result is a testament to the power of integral calculus in solving geometric problems, particularly in three dimensions. The process we followed highlights several key concepts and techniques that are essential for understanding and applying triple integrals.

Key takeaways from this exercise include:

1. Visualization is Crucial: The first step in solving any geometric problem is to visualize the shape and its boundaries. In this case, understanding that the solid was a tetrahedron in the first octant was essential for setting up the limits of integration. A clear mental picture or a sketch can greatly aid in this process.

2. Setting up the Limits of Integration: The most critical step in evaluating a triple integral is correctly determining the limits of integration. This involves understanding how the variables relate to each other within the given region. We systematically determined the limits for $z$, $y$, and $x$ by considering the planes that bounded the tetrahedron. The equation $x+y+z=1$ played a central role in defining these relationships.

3. Order of Integration: The order in which we integrate the variables ($x$, $y$, and $z$) matters. The limits of integration dictate the order. We integrated with respect to $z$ first, then $y$, and finally $x$, because the limits for $z$ were expressed in terms of $x$ and $y$, the limits for $y$ were expressed in terms of $x$, and the limits for $x$ were constant.

4. Step-by-Step Evaluation: Evaluating a triple integral involves performing the integration in a step-by-step manner, starting with the innermost integral and working outwards. This approach simplifies the problem and allows us to handle the integration one variable at a time. We carefully evaluated each integral, substituting the limits of integration at each step.

5. Substitution Techniques: Sometimes, a simple substitution can make the integration process easier. In our case, substituting $u = 1-x$ simplified the final integration step and allowed us to easily evaluate the integral. — Phoenix Craigslist Cars: Find Your Next Ride By Owner

This example of calculating the volume of a tetrahedron provides a valuable framework for tackling other three-dimensional volume problems. The principles of visualization, setting up limits of integration, and step-by-step evaluation are universally applicable. By mastering these techniques, you can confidently approach a wide range of problems in multivariable calculus and geometry. The application of triple integrals extends beyond simple volume calculations. They are used in physics to calculate moments of inertia and centers of mass, in engineering to analyze stress and strain in solids, and in computer graphics to render three-dimensional scenes. Understanding triple integrals is therefore a valuable skill for anyone working in these fields.

In summary, the calculation of the volume of the tetrahedron highlights the elegance and power of triple integrals. It demonstrates how calculus can be used to solve geometric problems and provides a foundation for understanding more complex applications of multivariable calculus. The key to success lies in careful visualization, precise setup of the integral, and methodical evaluation. With practice, these techniques become second nature, opening up a world of possibilities in mathematics, science, and engineering.