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Magazine Chapter 3: Problem 10
Find the volume \(V\) of the solid \(S\) bounded by the three coordinate planes, bounded above by the plane \(x+y+z=2,\) and bounded below by the plane \(z=x+y\).
Short Answer
Expert verified
The volume is \frac{8}{3}.
Step by step solution
01
- Understand the geometry
The solid is bounded by the coordinate planes (i.e., the planes where one of the coordinates is zero) and two additional planes: one at the top given by the equation \( x + y + z = 2 \) and one at the bottom given by \( z = x + y \).
02
- Identify the limits of integration
To find the volume, set up a triple integral. The limits of integration for each variable can be determined by the intersections of the planes. The region of integration in the xy-plane is determined by the projection of the intersection of the two planes, from \( x=0 \) to \( x=2 \) and \( 0 \) to \( 2-x \) for \( y \). The corresponding \( z \) limits range from \( z = x + y \) to \( z = 2 - x - y \).
03
- Set up the integral
Write the volume as a triple integral: \[ V = \int_{0}^{2} \int_{0}^{2-x} \int_{x+y}^{2-x-y} dz \, dy \, dx \].
04
- Evaluate the inner integral
First integrate with respect to \( z \): \[ \int_{x+y}^{2-x-y} dz = \left[ z \right]_{x+y}^{2-x-y} = (2-x-y) - (x+y) = 2-2x-2y. \].
05
- Evaluate the middle integral
Next, integrate the result with respect to \( y \): \[ \int_0^{2-x} (2-2x-2y) \, dy = 2y - 2xy - y^2 \Big|_0^{2-x} = 2(2-x) - 2x(2-x) - (2-x)^2 = 4 - 4x + x^2. \].
06
- Evaluate the outer integral
Lastly, integrate the result with respect to \( x \): \[ \int_0^2 (4-4x+x^2) \, dx = 4x - 2x^2 + \frac{1}{3}x^3 \Big|_0^2 = 8 - 8 + \frac{8}{3} = \frac{8}{3}. \].
07
Conclusion
So, the volume of the solid bounded by the given planes is \( \frac{8}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Triple Integral
A triple integral is a way to integrate a function of three variables over a three-dimensional region. Imagine stacking infinitely thin slices of a solid and summing their volumes. This process lets you find volumes of three-dimensional objects.
In our example, we calculate the volume of a solid bounded by different planes using a triple integral. The general form for a triple integral over a region \(\text{E}\) with density function \(\rho(x, y, z)\) is: \[ \iiint_E \rho(x, y, z) \,dV \. \]
We simplify the process by evaluating the integral in three stages:
In our example, we calculate the volume of a solid bounded by different planes using a triple integral. The general form for a triple integral over a region \(\text{E}\) with density function \(\rho(x, y, z)\) is: \[ \iiint_E \rho(x, y, z) \,dV \. \]
We simplify the process by evaluating the integral in three stages:
- First, integrating with respect to \(\text{z}\).
- Next, evaluating the result with respect to \(\text{y}\),
- Finally, integrating with respect to \(\text{x}\).
Limits of Integration
Establishing the limits of integration is foundational for solving a triple integral. These limits define the region over which we integrate.
In this problem, the integration limits come from the intersection of the planes:
In this problem, the integration limits come from the intersection of the planes:
- The region in the \(xy \)-plane is bounded from \(x = 0\) to \(x = 2\).
- \begin{align*} \int_{0}^{2}& \, dx, \int_{0}^{2-x}& \, dy, \int_{x+y}^{2-x-y}& \, dz. \end{align*}
Volume of Solid
The volume of a solid bounded by surfaces can be computed using triple integrals. In the given problem, we compute the volume of a solid enclosed between planes. The equation: \[ V = \int_{0}^{2} \int_{0}^{2-x} \int_{x+y}^{2-x-y} dz \, dy \, dx. \]
This triple integral is evaluated step-by-step:
This results in \(4 - 4x + x^2\). Finally, integrating this result with respect to \(\text{x}\):
Understanding the bounds and applying the triple integral step-by-step helps accurately determine the volume of the solid.
This triple integral is evaluated step-by-step:
- First, integrating with respect to \(\text{z}\) gives us: \[ \int_{x+y}^{2-x-y} dz. \]
- Evaluating this results in \(2 - 2x - 2y\).
- Next, integrating this with respect to \(\text{y}\) gives us: \[ \int_{0}^{2-x} (2-2x-2y)\, dy. \]
This results in \(4 - 4x + x^2\). Finally, integrating this result with respect to \(\text{x}\):
- \[ \int_{0}^{2} (4-4x+x^2) \,dx = \]
- \[\[\begin{align*} 4x - 2x^2 + \frac{1}{3}x^3 \end{align*}\]\].
Computing this yields the volume, which is \frac{8}{3}\.
Understanding the bounds and applying the triple integral step-by-step helps accurately determine the volume of the solid.
