Question

\(\int_{0}^{\pi} \int_{0}^{a x+1} r d r d \theta=\ldots \ldots\)

Short answer

\( \frac{1}{2} \pi (a x + 1)^2 \) is the result of the double integral.

Step-by-step solution

Understanding the Problem

The problem asks us to compute a double integral \( \int_{0}^{\pi} \int_{0}^{a x+1} r \, d r \, d \theta \). This involves integrating with respect to \( r \) first, within the limits from 0 to \( ax + 1 \), and then integrating with respect to \( \theta \), from 0 to \( \pi \).

Inner Integral Calculation

First, we tackle the inner integral \( \int_{0}^{a x+1} r \, d r \). This integral is straightforward: the integral of \( r \) with respect to \( r \) is \( \frac{r^2}{2} \). Applying the limits, we have:\[ \frac{(a x + 1)^2}{2} - \frac{0^2}{2} = \frac{(a x + 1)^2}{2}. \]

Setting Up the Outer Integral

Now that we have the result of the inner integral as \( \frac{(a x + 1)^2}{2} \), we substitute this result into the outer integral:\[ \int_{0}^{\pi} \frac{(a x + 1)^2}{2} \, d \theta. \] This simplifies to:\[ \frac{1}{2} \int_{0}^{\pi} (a x + 1)^2 \, d \theta. \]

Evaluating the Outer Integral

The integral with respect to \( \theta \) now involves a constant, \( (a x + 1)^2 \. \):\[ \int_{0}^{\pi} (a x + 1)^2 \, d \theta = (a x + 1)^2 \cdot \int_{0}^{\pi} \, d \theta. \] Evaluating the integral, \( \int_{0}^{\pi} \, d \theta = \pi - 0 = \pi \), the expression becomes:\[ (a x + 1)^2 \cdot \pi. \]

Final Result

Substituting back, the double integral thus evaluates to\[ \frac{1}{2} \cdot (a x + 1)^2 \cdot \pi. \]

Key concepts

Integral Calculus

Integral calculus is a fundamental concept in mathematics, focusing on the accumulation of quantities and the areas under and between curves. In simpler terms, it helps us calculate the total amount of something when we know its rate of change. Imagine having a function "r" that describes how a quantity varies. To find out how much this quantity changes overall, we use integration.
In our exercise, we see a classic problem involving integration, where the task is to compute a double integral, a more advanced form where two integrations are performed successively.

• The first integration is with respect to one variable, usually within specified limits.
• The result is then integrated again with respect to the second variable.

This approach helps in handling complex shapes and curves, essential for solving real-world applications like physics problems, engineering tasks, and other scientific inquiries.

Multiple Integrals

Multiple integrals extend the basic concept of a single integral to functions of two or more variables. They are useful in scenarios where quantities depend on multiple dimensions, like volume calculation and other multidimensional analyses.
The double integral serves as a particular case of multiple integrals. In the problem provided, the double integral is used to perform integration over a bounded area defined by two different variables, namely, "r" and "θ".

• The inner integral calculates the contribution of one variable over its range.
• The outer integral then aggregates these results across the range of the second variable.

By understanding multiple integrals, we gain powerful tools for exploring spaces beyond the one-dimensional, evaluating wide-ranging functions easily.

Mathematical Problem Solving

Mathematical problem-solving is an iterative process where complex problems are broken down into simpler steps. The goal is to tackle each piece methodically until the whole problem is resolved. In the context of our double integral, solving the problem requires an understanding of:

• Identifying the limits of integration and their logical order.
• Computing the inner integral, which affects the subsequent outer integral surrounding it.
• Correctly substituting results from one step into the next to maintain order and achieve an accurate final answer.

Mathematical problem-solving is akin to piecing together a puzzle, where each part must fit accurately with others to complete the entire picture. It's a vital skill not only for integration tasks but for any analytical challenge encountered in academics or professional life.

Integration Techniques

Integration techniques are essential methods used to evaluate integrals. Each technique is selected based on the nature of the function to be integrated. For the exercise at hand, the steps can be facilitated using basic yet effective techniques:

• Substitution: Identifying intermediate expressions to simplify integrals.
• Separation: Tackling complex integrals one variable at a time.

In practical terms, the integration begins with computing the simple polynomial function from the inner integral, which leads to straightforward substitution or application of standard formulas. Following, the result becomes a constant alongside the outer integral, which is then resolved through elementary integration rules.
These techniques not only speed up calculations but also ensure accuracy, enabling us to solve intricate mathematical models efficiently and comprehensibly.