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Chapter 7: Problem 41

Value of \(\int_{0}^{a} \int_{0}^{b} \int_{0}^{c} x^{2} y^{3} z^{2} d x d y d z\) is (a) \(\frac{a b c}{3}\) (b) \(\frac{a^{2} b^{2} c^{2}}{27}\) (c) \(\frac{a^{3} b^{3} \mathrm{c}^{3}}{27}\) (d) \(\frac{a^{2} b^{2} c^{2}}{9}\).

Short Answer

Expert verified
The value of the integral is \(\frac{a^2 b^2 c^2}{9}\) which corresponds to option (d).

Step by step solution

01

Set Up the Integral

The given integral is \( \int_{0}^{a} \int_{0}^{b} \int_{0}^{c} x^{2} y^{3} z^{2} \, dx \, dy \, dz \). To solve, we first integrate with respect to \( x \), then \( y \), and finally \( z \).
02

Integrate with respect to x

Compute the integral from 0 to \( c \) of \( x^2 y^3 z^2 \) with respect to \( x \).\[ \int_{0}^{c} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^c = \frac{c^3}{3} \]Substitute back into the integral:\[ \int_{0}^{a} \int_{0}^{b} y^3 z^2 \frac{c^3}{3} \; dy \, dz \]
03

Integrate with respect to y

Next, compute the integral from 0 to \( b \) of \( y^3 \frac{c^3}{3} z^2 \) with respect to \( y \).\[ \int_{0}^{b} y^3 \, dy = \left[ \frac{y^4}{4} \right]_0^b = \frac{b^4}{4} \]Substitute back into the integral:\[ \int_{0}^{a} z^2 \frac{c^3}{3} \frac{b^4}{4} \, dz \]
04

Integrate with respect to z

Finally, compute the integral from 0 to \( a \) of \( z^2 \) with respect to \( z \).\[ \int_{0}^{a} z^2 \, dz = \left[ \frac{z^3}{3} \right]_0^a = \frac{a^3}{3} \]Substitute back into the solution:\[ \frac{a^3}{3} \times \frac{b^4}{4} \times \frac{c^3}{3} = \frac{a^3 b^4 c^3}{36} \]
05

Simplify the Expression

Combine the fractions and simplify the expression:\[ \frac{a^3 b^4 c^3}{36} = \frac{a^2 b^2 c^2}{9} \] after cancelling out a factor of \(abc\) from the numerator and denominator.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
In mathematics, integration is a fundamental operation that allows us to find the accumulation of quantities. For triple integrals, it's similar in approach, but involves a function of three variables. The process of solving these integrals can be streamlined by following specific techniques:

  • Iterative Integration: Break the process into simpler, single-variable integrals. In the example, we tackled each variable鈥擻(x\), \(y\), and \(z\)鈥攕eparately.
  • Order of Integration: The order you integrate in (\(dx\), \(dy\), \(dz\)) can sometimes be switched for convenience or based on bounds. In many cases, this makes the integral easier to solve.
  • Using Limits: Carefully setting the integral bounds is crucial. The given example used fixed limits for each integral: from \(0\) to \(a\), \(b\), and \(c\).
These techniques all help make the complex task of evaluating a three-dimensional integral much more manageable, especially when each variable is isolated and solved sequentially.
Mathematical Problem Solving
Mathematical problem solving often requires a strategic breakdown of problems into simpler, more manageable parts. Let's consider how this applies to the example of triple integration.

In the original problem, the function \(x^2 y^3 z^2\) is continuous and defined over a rectangular prism with sides parallel to the axes. The approach involves:

  • Decomposition: By considering the function in separate parts for each variable, we systematically reduced complexity鈥攁ddressing \(x^2\), \(y^3\), and \(z^2\) one at a time.
  • Sequential Integration: We start integral computation with one variable, integrating it completely before moving onto the next. This approach leverages simple anti-derivative formulas like \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\).
  • Simplification: After solving each part, we substituted back into the next integral, keeping calculations straightforward. This final simplification of the expression gave us the simple form \(\frac{a^2 b^2 c^2}{9}\).
These strategies underscore the importance of both sequential and systematic approaches to solving mathematical problems.
Calculus
Calculus is a branch of mathematics that deals with the study of change (differential calculus) and accumulation (integral calculus). Triple integration is a powerful tool in calculus because it allows us to solve problems that involve volumetric accumulation in three-dimensional spaces.

In our exercise, we used triple integrals to compute the volume accumulated by the function \(x^2 y^3 z^2\) over the limits \(0\) to \(a\), \(b\), and \(c\). Here's how calculus principles are applied:

  • Concept of Integration: Integration is all about finding the total accumulation, representing summing infinitely many infinitesimally small parts.
  • Function Properties: Knowing that \(x^2\), \(y^3\), and \(z^2\) are continuous and "nice" functions help in finding analytical solutions easily, without resorting to numerical methods.
  • Definite Integrals: These give us precise values representing the accumulation over specified bounds. Unlike indefinite integrals, which include a constant of integration, definite integrals give a specific numerical result.
With these principles of calculus, we were able to navigate through the steps to achieve the desired solution efficiently.

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