Question

Evaluate the triple integrals. $$\int_{x=1}^{2} \int_{y=x}^{2 x} \int_{z=0}^{y-x} d z d y d x$$

Short answer

\[ \frac{7}{6} \].

Step-by-step solution

Understand the Integral

The given integral \( \int_{x=1}^{2} \int_{y=x}^{2x} \int_{z=0}^{y-x} dz \, dy \, dx \) is a triple integral. It involves integrating with respect to \(z\), \(y\), and \(x\) in that specific order.

Integrate with Respect to \(z\)

First, integrate the innermost integral \( \int_{z=0}^{y-x} dz \). Since the integrand is 1, this becomes: \int_{z=0}^{y-x} dz = z \bigg|_{0}^{y-x} = y-x.

Substitute and Integrate with Respect to \(y\)

Next, substitute the result of the inner integral back into the second integral: \int_{x=1}^{2} \int_{y=x}^{2x} (y-x) dy \,dx. Now, we integrate with respect to \(y\). The integral becomes: \int_{y=x}^{2x} (y-x) dy = \left[ \frac{(y-x)^2}{2} \right]_{y=x}^{y=2x}.

Evaluate the Second Integral

Evaluate the integral from \(y=x\) to \(y=2x\): \frac{(2x-x)^2}{2} - \frac{(x-x)^2}{2} = \frac{x^2}{2}.

Integrate with Respect to \(x\)

Substitute the result back into the outermost integral: \int_{x=1}^{2} \frac{x^2}{2} dx. Now, integrate with respect to \(x\). The integral becomes: \int_{1}^{2} \frac{x^2}{2} dx = \frac{1}{2} \left[ \frac{x^3}{3} \right]_{1}^{2}.

Evaluate the Final Integral

Evaluate \[\frac{1}{2} \left( \frac{2^3}{3} - \frac{1^3}{3} \right) = \frac{1}{2} \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{1}{2} \times \frac{7}{3} = \frac{7}{6} \]. Hence, the final value of the triple integral is \[\frac{7}{6} \].

Key concepts

integration

Integration is a fundamental concept in calculus. It helps us find areas under curves, volumes, and other quantities that accumulate over a range. In our problem, we are dealing with a **triple integral**. To understand integration better:

- **Integrand**: This is what you're integrating. In this case, it's originally just 1 within the bounds.
- **Bounds of Integration**: These tell us where to start and stop integrating. For the innermost integral, upper and lower limits depend on the values of other variables.
The process involves integrating step-by-step over each variable from the innermost to the outermost. Each level of integration helps break down a complex problem into manageable parts.

multivariable calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of more than one variable. Here, instead of finding an area or a curve, we’re finding a volume in three-dimensional space by evaluating a triple integral. Let’s explore some key points:
- **Multiple Dimensions**: With triple integrals, we’re working with three variables, typically x, y, and z. Each integration 'removes' a dimension.
- **Bounded Region**: The region of integration is determined by the given limits for x, y, and z. In our problem, we integrate within a specified 3D region.
- **Iterative Integration**: By performing integration step-by-step, we simplify the problem. We start with the innermost integral and work outward. This helps in treating each variable separately, reducing the chances of errors.
Multivariable calculus is essential in fields like physics, engineering, and economics, where problems often involve multiple changing quantities.

step-by-step solution

A step-by-step solution helps break down complex math problems into simple, understandable steps. Let’s revisit the given triple integral:
1. **Identify the Order**: The integral is \int_{x=1}^{2} \int_{y=x}^{2x} \int_{z=0}^{y-x} dz \, dy \, dx. This means: integrate with respect to z first, then y, and finally x.
2. **First Integration**: Integrate \int_{z=0}^{y-x} dz. Since the integrand is 1, this simplifies to **y-x**.
3. **Next Integration**: Substitute and integrate \int_{y=x}^{2x} (y-x) dy. Substituting the bounds, we find \[ \frac{(2x-x)^2}{2} - \frac{(x-x)^2}{2} = \frac{x^2}{2} \].
4. **Final Integration**: Substitute back and integrate \int_{x=1}^{2} \frac{x^2}{2} dx. This evaluates to \[ \frac{1}{2} \times \frac{7}{3} = \frac{7}{6} \].
Breaking it down not only clarifies each step but also helps uncover any errors. Following step-by-step solutions is key in mastering calculus.