Question

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer. $$\int_{y=-2}^{1} \int_{x=1}^{2} 8 x y d x d y$$

Short answer

The value of the double integral is \-18.

Step-by-step solution

Identify the region of integration

The integral is given over the region with bounds for x from 1 to 2 and bounds for y from -2 to 1. This implies we are integrating over a rectangular region in the xy-plane.

Set up the double integral

Express the double integral with the given bounds: \[ \int_{y=-2}^{1} \int_{x=1}^{2} 8xy \, dx \, dy \]

Integrate with respect to x

First, integrate the inner integral with respect to x: \( \int_{x=1}^{2} 8xy \, dx \)Treating y as a constant, integrate: \[ \int_{1}^{2} 8xy \, dx = y \int_{1}^{2} 8x \, dx = 8y \left [ \frac{x^{2}}{2} \, \right ]_{1}^{2} \]Substitute the limits of integration and simplify: \[ 8y \left ( \frac{2^{2}}{2} - \frac{1^{2}}{2} \right ) = 8y \left ( 2 - \frac{1}{2} \right ) = 8y \left ( \frac{3}{2} \right ) = 12y \]

Integrate with respect to y

Now, integrate the result with respect to y: \[ \int_{-2}^{1} 12y \, dy \]Integrate: \[ 12 \int_{-2}^{1} y \, dy = 12 \left [ \frac{y^{2}}{2} \right ]_{-2}^{1} \]Substitute the limits of integration and simplify: \[ 12 \left ( \frac{1^{2}}{2} - \frac{(-2)^{2}}{2} \right ) = 12 \left ( \frac{1}{2} - 2 \right ) = 12 \left ( -\frac{3}{2} \right ) = -18 \]

Check results by computer (optional)

Verify the result using a computational tool such as a graphing calculator, a computer algebra system, or an online integrator to ensure the correctness of the manual calculations.

Key concepts

Region of Integration

The region of integration specifies where we are integrating in the xy-plane. For the given integral, the bounds are:

• x ranges from 1 to 2
• y ranges from -2 to 1

This means our region of interest is a rectangle in the xy-plane. Think of this like a box drawn on a graph where x goes horizontally from 1 to 2 and y goes vertically from -2 to 1. This ‘box’ encloses the area over which we will evaluate our function.

Inner Integral

The inner integral refers to the first integration operation you perform. In this problem, the inner integral is with respect to x: \( \int_{x=1}^{2} 8xy \, dx \)

Here, we treat y as a constant while integrating with respect to x. Think of it as slicing the region vertically and adding up those slices. First, we find the antiderivative of 8xy with respect to x:
\( 8y \int_{1}^{2} x \, dx \)
The integral of x with respect to x is \( \frac{x^2}{2} \), so we get:
\( 8y \left [ \frac{x^2}{2} \right ]_{1}^{2} \)

Outer Integral

After evaluating the inner integral, the outer integral is the next task. This step involves integrating with respect to the remaining variable, y. Using our result from the inner integral (12y), we now need to integrate with respect to y over the given bounds:
\( \int_{y=-2}^{1} 12y \, dy \)

Here, we integrate the simpler expression 12y. It’s like summing up the contributions from each horizontal strip of the rectangle. The integral of y with respect to y is:
\( \frac{y^2}{2} \), which upon evaluating from -2 to 1 gives:
\( 12 \left [ \frac{y^2}{2} \right ]_{-2}^{1} \)

Bounds of Integration

The bounds of integration determine the limits within which you perform your integration. For a double integral, there are two sets of bounds: one for each variable.
In our example:

• The bounds for x (inner integral) are from 1 to 2
• The bounds for y (outer integral) are from -2 to 1

These bounds define where we start and stop our integration for each variable. Properly setting bounds is crucial because it precisely defines the area or volume over which we evaluate the integral. Always pay attention to the problem's bounds, as they directly impact the solution.