Question

Evaluate (a) \(\int_{R}(x-2 y+3 z) d(x, y, z) ; \quad R=[-2,0] \times[2,5] \times[-3,2]\) (b) \(\int_{R} e^{-x^{2}-y^{2}} \sin x \sin z d(x, y, z) ; \quad R=[-1,1] \times[0,2] \times[0, \pi / 2]\) (c) \(\int_{R}(x y+2 x z+y z) d(x, y, z) ; \quad R=[-1,1] \times[0,1] \times[-1,1]\) (d) \(\int_{R} x^{2} y^{3} z e^{x y^{2} z^{2}} d(x, y, z) ; \quad R=[0,1] \times[0,1] \times[0,1]\)

Short answer

In summary, the results for each part of the exercise are as follows: (a) The value of the triple integral is -153. (b) The value of the triple integral is 0, as the integral of an odd function over a symmetric interval is zero. (c) The value of the triple integral is 0. (d) The value of the triple integral is \(\frac{1}{12}\).

Step-by-step solution

Identify the integrand and the limits of integration

The integrand is \((x - 2y + 3z)\) and the region of integration is \(R=[-2,0] \times[2,5] \times[-3,2]\).

Integrate with respect to x

\(\int_{-2}^{0}(x - 2y + 3z) dx = [\frac{1}{2}x^2 - 2xy + 3xz]_{-2}^0\).

Integrate with respect to y

\(\int_{2}^{5} [\frac{1}{2}x^2 - 2xy + 3xz] dy = [\frac{1}{2}x^2y - xy^2 + 3xyz]_{2}^5\).

Integrate with respect to z

\(\int_{-3}^{2}[\frac{1}{2}x^2y - xy^2 + 3xyz] dz = [\frac{1}{2}x^2yz - xyz^2 + \frac{3}{2}xyz^2]_{-3}^2\).

Calculate the final value

Substitute the limits and simplify to get the final value of the integral. The result is -153. (b)

Identify the integrand and the limits of integration

The integrand is \(e^{-x^2-y^2} \sin x \sin z\) and the region of integration is \(R=[-1,1] \times[0,2] \times[0, \frac{\pi}{2}]\).

Integrate with respect to x

\(\int_{-1}^{1}e^{-x^2-y^2} \sin x \sin z dx = \frac{1}{2}e^{-y^2} (\sin z) \int_{-1}^{1}e^{-x^2} (\sin x) dx\). To evaluate the integral, we use the property of odd functions, that is, the integral of an odd function over a symmetric interval is zero. Since \(\int_{-1}^{1}e^{-x^2} (\sin x) dx\) is an odd function, the result is 0. (c)

Identify the integrand and the limits of integration

The integrand is \((xy + 2xz + yz)\) and the region of integration is \(R=[-1,1] \times[0,1] \times[-1,1]\).

Integrate with respect to x

\(\int_{-1}^{1}(xy + 2xz + yz) dx = [\frac{1}{2}x^2y + x^2z + xyz]_{-1}^1\).

Integrate with respect to y

\(\int_{0}^{1}[\frac{1}{2}x^2y + x^2z + xyz] dy = [\frac{1}{4}x^2y^2 + xy^2z + \frac{1}{2}xyz^2]_{0}^1\).

Integrate with respect to z

\(\int_{-1}^{1}[\frac{1}{4}x^2z^2 + xz^2 + \frac{1}{2}xz^3] dz = [\frac{1}{12}x^2z^3 + \frac{1}{3}xz^3 + \frac{1}{8}xz^4]_{-1}^1\).

Calculate the final value

Substitute the limits and simplify to get the final value of the integral. The result is 0. (d)

Identify the integrand and the limits of integration

The integrand is \(x^2y^3ze^{x y^2 z^2}\) and the region of integration is \(R=[0,1] \times[0,1] \times[0,1]\).

Integrate with respect to x

\(\int_{0}^{1}x^2y^3ze^{x y^2 z^2} dx = \int_{0}^{1}x^2y^3ze^{x y^2 z^2} dx\). To evaluate the integral, apply integration by parts.

Integrate with respect to y

The integral is now \(\int_{0}^{1}(\frac{1}{6}y^3z^4e^{y^2 z^2})\). We will apply the substitution method to solve for the integral.

Integrate with respect to z

The integral is now \(\int_{0}^{1}(\frac{1}{12}z^4e^{z^2})\). Apply the substitution method once again to solve for the integral.

Calculate the final value

Substitute the limits and simplify to get the final value of the integral. The result is \(\frac{1}{12}\).

Key concepts

Triple Integrals

Triple integrals extend the idea of single and double integrals to three dimensions. They are used to calculate the volume under a surface in three-dimensional space, or to integrate functions over a three-dimensional region. In the exercise above, we are looking at triple integrals over a rectangular prism, defined by specific limits for each variable: \(x\), \(y\), and \(z\). Each integral resolves within these bounds, leading to the cumulative evaluation across three axes sequentially.
Essentially, we begin integrating with respect to one variable, while treating the others as constants. For instance, while performing the integral with respect to \(x\) first, we consider \(y\) and \(z\) as constants. This approach is critical because:

• It simplifies the problem, breaking it down into smaller, manageable parts.
• Preserves the dependencies across dimensions, which is vital for accurate computation.

Understanding the fundamental concept of triple integrals helps grasp how multi-dimensional calculus applies in real-world applications, such as physics and engineering.

Integration Techniques

In multivariable calculus, various integration techniques can be employed to evaluate complex integrals. The basic methods include standard indefinite or definite integration, substitution, and integration by parts. In our example:- **Substitution Method**: This involves replacing a particular expression in the integral with a simpler variable to make integration more straightforward. It is particularly useful when dealing with complex functions inside the integral. - **Integration by Parts**: This technique, derived from the product rule for differentiation, is beneficial when integrating products of functions, like the one in problem (d). It follows the formula: \[ \int u \, dv = uv - \int v \, du \] Here, choosing \(u\) and \(dv\) carefully can simplify the integral significantly.Using these techniques effectively requires practice and an understanding of when each method is best suited for the integral at hand. Often, the choice of technique will depend on the complexity and form of the function being integrated, as well as the region over which it is integrated.

Regions of Integration

Regions of integration in multivariable calculus define the space over which we integrate our function. In the cases provided, these regions are defined as rectangular prisms by boundaries of each variable. For example, in problem (a), the bounds are given as \([-2, 0] \times [2, 5] \times [-3, 2]\).
This means:

• The variable \(x\) ranges from -2 to 0.
• \(y\) ranges from 2 to 5.
• \(z\) ranges from -3 to 2.

These bounds define a finite region in three-dimensional space where the function will be integrated. Understanding these boundaries is crucial because they determine the limits of each individual integral step. Moreover, the choice of order for integrating over these dimensions (such as \(x\) followed by \(y\), and then \(z\)) can sometimes simplify computation, particularly if certain dimensions have more complex interaction terms.Identifying and handling multiple types of regions of integration efficiently is an important skill in solving triple integrals in various applications, making calculus a powerful tool in mathematical modeling and analysis.