Question

\(\iint x^{2} y^{4}\) dxdy over the rectangle \(0 \leq x \leq 1\) and \(0 \leq y \leq 3\) is

Short answer

The integral evaluates to \(\frac{81}{5}\).

Step-by-step solution

Understanding the Problem

We need to evaluate the double integral \( \iint x^{2} y^{4} \, dx \, dy \) over the rectangular region \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 3 \). This means we first integrate with respect to \(x\) between 0 and 1, and then integrate the result with respect to \(y\) between 0 and 3.

Set Up the Integral

Write the integral in the form of iterated integrals: \[ \int_{0}^{3} \int_{0}^{1} x^{2} y^{4} \, dx \, dy \] This means we will first perform the integration with respect to \(x\) and afterwards with respect to \(y\).

Integrate with Respect to x

Compute the integral with respect to \(x\): \[ \int_{0}^{1} x^{2} y^{4} \, dx \] Treat \(y\) as a constant, so the integral becomes \( y^4 \int_{0}^{1} x^{2} \, dx \). The antiderivative of \(x^2\) is \(\frac{x^3}{3}\), so evaluate from 0 to 1: \[ y^{4} \left[ \frac{x^3}{3} \right]_{0}^{1} = y^{4} \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = \frac{y^4}{3} \]

Integrate with Respect to y

Now integrate the result with respect to \(y\): \[ \int_{0}^{3} \frac{y^4}{3} \, dy \] The antiderivative of \(y^4\) is \(\frac{y^5}{5}\), so evaluate from 0 to 3: \[ \frac{1}{3} \left[ \frac{y^5}{5} \right]_{0}^{3} = \frac{1}{3} \left( \frac{3^5}{5} - \frac{0^5}{5} \right) = \frac{1}{3} \times \frac{243}{5} = \frac{243}{15} = \frac{81}{5} \]

Final Answer

The value of the double integral \( \iint x^{2} y^{4} \, dx \, dy \) over the given rectangle is \( \frac{81}{5} \).

Key concepts

Iterated Integrals

Iterated integrals are a critical part of evaluating double integrals. When we have a double integral, we essentially are performing integration twice, once for each variable. In the given problem, the integral is evaluated first with respect to the variable \( x \) and then \( y \).
The idea here is simple: start from the inside out. Integrate \( x^2 y^4 \) treating \( y \) as a constant when integrating with respect to \( x \). Once we have a simplified form, we proceed to integrate with respect to \( y \). This structured approach, where one integration is nested within another, is termed as iterated integration.
Iterated integrals help in breaking down multi-variable problems into manageable single-variable calculus operations, greatly simplifying the process of finding the area or volume over a specified region.

Antiderivative

Antiderivatives play a central role in solving integrals. Essentially, an antiderivative of a function is another function whose derivative returns the original function. When evaluating the given double integral \( \iint x^2 y^4 \ dx \ dy \), we first find the antiderivative of \( x^2 \), since \( y \) is treated as a constant.
For \( x^2 \), the antiderivative is \( \frac{x^3}{3} \). Similarly, when we then integrate with respect to \( y \), the antiderivative of \( y^4 \) is \( \frac{y^5}{5} \).
Finding these antiderivatives allows us to simplify and evaluate the expressions within set bounds, ultimately leading us to the solution of the problem. For all variables, understanding how to determine the antiderivative is essential for the successful calculation of any integral.

Rectangular Region

The rectangular region is integral to setting up a double integral as it defines the limits of integration. In our case, the rectangle is defined by \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 3 \).
These limits are critical because they determine the boundaries over which we perform our integrations. For instance, we integrate \( x \) from 0 to 1, and \( y \) from 0 to 3. This rectangle provides a structured way to tackle multidimensional integrals and is often represented in both mathematical notation and visual diagrams.
Understanding and properly setting these limits is essential because incorrect bounds lead to wrong calculations and answers. The rectangular region aids in visualizing the specific area being analyzed, grounding abstract math in geometric reasonings.

Integration Techniques

A variety of integration techniques are used when solving integrals, making these problems some of the most diverse in calculus. For double integrals, both substitution methods and iterated integrals are often used. In our exercise, the primary method employed is iterated integration.
When evaluating \( \iint x^2 y^4 \ dx \ dy \), integration by parts isn’t necessary, and neither are substitution methods, given the straightforward polynomial form. The focus was on handling the integral one variable at a time, simplifying each step by finding antiderivatives directly.
Techniques such as recognizing constants during integration—like treating \( y \) as a constant during the \( x \) integration—simplify the process. Mastering these techniques allows students to tackle a wide array of integral problems, particularly those involving more than one dimension.