91Ó°ÊÓ

Integration limits define the boundaries over which we perform the integration. In the case of a double integral, we have limits for both variables involved.
For this exercise, the region \(D\) is defined by the limits \(-1 \leq y \leq 1\) and \(y^2 - 1 \leq x \leq \sqrt{1-y^2}\). This means:

• The variable \(y\) is the outer variable, indicating that it spans from \(-1\) to \(1\). This creates a symmetrical range.
• The variable \(x\) is the inner variable, and its limits are dependent on \(y\). The lower limit is \(y^2 - 1\) and the upper limit is \(\sqrt{1-y^2}\), showing that \(x\)'s range varies with \(y\).

Setting these limits accurately is crucial for correctly evaluating the double integral.

A definite integral provides the total accumulation of a function over a specific interval, producing a fixed number as a result. It offers the total area under the curve defined by the function between the specified points.
For a double integral, as seen in this exercise, it's like stacking sheets of area together over a region within a plane.
When we set up our integral as \(\int_{-1}^{1} \int_{y^2 - 1}^{\sqrt{1-y^2}} x y \; dx \, dy\), we're calculating the total volume below the surface described by \(f(x, y) = xy\) and above the region \(D\).
The limits are crucial in determining where our accumulation stops and starts, converting potentially infinite calculations into a precise value.

Integrating with respect to one variable at a time in a double integral simplifies the computation process.
When integrating with respect to \(x\), we treat \(y\) as a constant. The integral \( \int (xy) \, dx\) becomes straightforward since \(y\) is just a coefficient. The result is \( \frac{x^2 y}{2} + C\) because the integral of \(x\) is \(\frac{x^2}{2}\).
After obtaining this expression, it's crucial to evaluate it using the integration limits for \(x\), which are \(x = y^2 - 1\) and \(x = \sqrt{1-y^2}\). By substituting these limits, we get:

• \(\frac{(\sqrt{1-y^2})^2 y}{2} - \frac{(y^2 - 1)^2 y}{2}\)
• Which simplifies to: \(\frac{y (1-y^2)}{2} - \frac{y^5 - 2y^3 + y}{2}\)

Effectively, we've broken down one layer of the problem.

Once the integral with respect to \(x\) is calculated and simplified, the next step is to perform integration with respect to \(y\).
For this exercise, after simplifying the expression, we needed to evaluate \(\int_{-1}^{1} \left( \frac{y^3}{2} - \frac{y^5}{2} \right) \, dy\). Here, \(\int y^n \, dy\) gives \( \frac{y^{n+1}}{n+1}\).
Applying this to our terms separately:

• \( \frac{y^4}{8} \) and \( -\frac{y^6}{12} \).

The limits \(y = -1\) and \(y = 1\) are then inserted to evaluate the definite integral. After substituting these, the final result of 0 illustrates that the areas computed above and below the \(x\)-axis cancel each other out, perfectly highlighting the beauty of definite integrals.