91Ó°ÊÓ

In the context of double integrals, the **region of integration** is crucial as it defines the area over which you perform the integration. Let's consider our specific problem where the region \( R \) is defined by the set \( \{ (x, y) | 0 \leq x \leq 3, 0 \leq y \leq 1 \} \). This describes a rectangle in the xy-plane.

The region is bounded by:

• **x-values**: ranging from 0 to 3
• **y-values**: ranging from 0 to 1

Understanding this rectangular region helps in setting the limits of integration correctly, which is a key first step in evaluating a double integral.

When sketching or visualizing, picture a simple rectangle, laying flat in the first quadrant of a coordinate plane, stretching between these boundaries. This visual aid can make it easier to grasp how the integration is applied over the specified range.

The concept of an **antiderivative** refers to finding a function that reverses the process of differentiation. When working with integrals, especially in a double integral context, you obtain the antiderivative of the function being integrated.

In the given problem, we start with the function \(6x^2y^3 - 5y^4\) and integrate with respect to \(y\) first:

• The antiderivative of \(6x^2y^3\) is \(\frac{6x^2y^4}{4}\) which simplifies to \(\frac{3x^2y^4}{2}\).
• The antiderivative of \(-5y^4\) is \(-\frac{5y^5}{5}\), simplifying to \(-y^5\).

These antiderivatives are then evaluated over the bounds of the integration to find the specific area under the curve, or volume in the case of double integrals.

The process of **integration with respect to variables** in double integrals involves calculating iteratively. You integrate one variable at a time, holding all others constant until each variable is integrated over its specified boundary.

With our integral, \(\int_{0}^{3} \int_{0}^{1} (6x^2y^3 - 5y^4) \, dy \, dx\), the idea is to:

• Integrate with respect to \(y\) first while treating \(x\) as a constant. This is done across its limits from 0 to 1.
• Then use the result of this integration in the outer integral, applying the antiderivative with respect to \(x\), across its limits from 0 to 3.

Complex double integrals become much more manageable by isolating variables in this sequential manner.

After determining the antiderivatives using their respective variables, the next step is the **evaluation of integrals**. This critical step involves substituting the boundary values into the antiderivative expression, which results in a numerical value representing the solution.

In our double integral example:

• First evaluating the \(y\)-antiderivative between 0 and 1 yields \( \frac{3x^2}{2} - 1 \).
• Then it becomes the function to integrate with respect to \(x\).
• The \(x\)-antiderivative is evaluated from 0 to 3, resulting in \(\frac{27}{2} - 3\), simplifying to \(\frac{21}{2}\).

Thus, completing the double integral provides a deeper understanding of the volume under the surface defined by the function \(6x^2y^3 - 5y^4\) over the region \(R\).

Paying careful attention to each step not only ensures accurate results but also reinforces the underlying mathematical principles.