Problem 34 - Thomas Calculus

Question

In Exercises \(31-40,\) sketch the region of integration, reverse the order of integration, and evaluate the integral. $$ \int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{e^8 - 1}{4}\).

Step 1: Sketch the Region of Integration

First, identify the bounds for the variables. For \(x\), the limits are from \(0\) to \(2\). For \(y\), the limits are from \(0\) to \(4-x^2\). This describes the region bounded by \(x = 0\), \(x = 2\), \(y = 0\), and \(y = 4 - x^2\). The curve \(y = 4 - x^2\) represents a downward-opening parabola intersecting the \(y\)-axis at \(y = 4\). The region in question is the area under this parabola, above the \(x\)-axis, for \(0 \leq x \leq 2\).

Step 2: Determine New Integration Bounds

To reverse the order of integration, solve \(y = 4 - x^2\) for \(x\). Rearrange to get \(x = \sqrt{4-y}\). For different values of \(y\), \(x\) will now vary from \(0\) to \(\sqrt{4-y}\) and \(y\) will range from \(0\) to \(4\). Thus, the double integral can be rewritten as \(\int_{0}^{4} \int_{0}^{\sqrt{4-y}} \frac{x e^{2y}}{4 - y} \, dx \, dy\).

Step 3: Evaluate the Inner Integral

Evaluate \(\int_{0}^{\sqrt{4-y}} \frac{x e^{2y}}{4-y} \, dx\). Since \(e^{2y} \) is constant with respect to \(x\), it can be factored out: \(\frac{e^{2y}}{4-y} \int_{0}^{\sqrt{4-y}} x \, dx\). The antiderivative of \(x\) is \(\frac{x^2}{2}\). Evaluate it from \(0\) to \(\sqrt{4-y}\), obtaining \(\frac{(4-y)e^{2y}}{2(4-y)} = \frac{e^{2y}}{2}\).

Step 4: Evaluate the Outer Integral

Now, integrate with respect to \(y\): \(\int_{0}^{4} \frac{e^{2y}}{2} \, dy\). The antiderivative of \(\frac{e^{2y}}{2}\) is \(\frac{e^{2y}}{4}\). Evaluate from \(0\) to \(4\), resulting in \(\frac{e^8}{4} - \frac{1}{4}\).

Step 5: Finalize the Solution

Combine the results from previous steps to find the value of the integral. The final answer for the integral is:\(\frac{e^8 - 1}{4}\). This completes the evaluation of the integral by reversing the order of integration.

Key Concepts

Order of Integration

When dealing with double integrals, the order of integration refers to the sequence in which integration is performed over the variables. This sequence can be either \( dy \) first, then \( dx \), or vice versa. Choosing the correct order of integration can simplify calculations significantly.

In the original exercise, the given order of integration is \( \int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} \, dy \, dx \). Here, we integrate with respect to \( y \) first for a given \( x \), followed by \( x \).

However, to simplify, we reversed the order to \( \int_{0}^{4} \int_{0}^{\sqrt{4-y}} \frac{x e^{2y}}{4 - y} \, dx \, dy \). By doing this, we made the computations easier since the bounding region was simpler to describe with this new order. Reversing the integration order often involves solving for one variable in terms of another from the given bounds.

Region of Integration

The region of integration defines the set of points over which the integration is performed. It usually involves sketching a graph or visualizing the area defined by the limits of integration.

In the exercise, the region is outlined by the inequalities: \(0 \leq x \leq 2\) and \(0 \leq y \leq 4 - x^2\). This describes a parabolic region in the first quadrant. The curve \(y = 4 - x^2\) limits the top of the region and is a downward-opening parabola.

Sketching helps in understanding the shape and constraints of this region. It acts as a visual guide when setting up and potentially reversing the order of integration. A change in integration order requires redefining this region boundary, hence sketching helps confirm your new bounds are accurate.

Definite Integral

A definite integral calculates the accumulated value of a function across a specified interval. With double integrals, we compute over a two-dimensional region, essentially finding the volume under a surface.

In our exercise, the definite integral \(\int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} dy dx\) has been evaluated by also reversing the order. This allowed us to manage the integral more comfortably. The definite integral, by its nature, provides a precise numerical result, in this case, \(\frac{e^8 - 1}{4}\).

Evaluating definite integrals effectively means managing both the integration process and accurately handling limits to ensure the region specified is correctly integrated.

Parabolic Region

The specific shape known as a parabolic region is defined by the equation of a parabola, a common form being \( y = ax^2 + bx + c \). In this context, we have the parabola \( y = 4 - x^2 \), which curves downward.

This specific parabola intersects the y-axis at \( y = 4 \) and is symmetrical around the y-axis, forming a boundary for our region of integration. Visually, the region lies below this curve and above the x-axis from \( x = 0 \) to \( x = 2 \).

Recognizing and sketching the shape of such parabolic regions is not only helpful but often necessary when adjusting the order of integration. By comprehending its geometry, one accurately determines the new boundaries needed when changing integration order, simplifying the calculation process.