Problem 2 - Thomas Calculus

Question

In Exercises \(1-10,\) sketch the region of integration and evaluate the integral. $$ \int_{0}^{3} \int_{-2}^{0}\left(x^{2} y-2 x y\right) d y d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-9\).

Step 1: Understand the region of integration

The given integral is a double integral with the limits of integration for \(y\) from \(-2\) to \(0\), and for \(x\) from \(0\) to \(3\). This defines a rectangular region in the xy-plane between the lines \(x=0\), \(x=3\), \(y=-2\), and \(y=0\). Sketching this on the xy-plane, we have a rectangle with vertices at \((0, -2), (3, -2), (3, 0)\), and \((0, 0)\).

Step 2: Integrate with respect to y

First, perform the inner integral with respect to \(y\). The inner integral is \(\int_{-2}^{0} (x^2 y - 2xy) \, dy\).\Compute: \[ \int (x^2 y - 2xy) \, dy = x^2 \cdot \frac{y^2}{2} - 2x \cdot \frac{y^2}{2}\Bigg|_{y=-2}^{y=0} = (x^2 - x)\left( \frac{y^2}{2} \right)\Bigg|_{y=-2}^{y=0} \]Evaluate the expression:\[ (x^2 - x) \left( \frac{(0)^2}{2} - \frac{(-2)^2}{2} \right) = (x^2 - x)(0 - 2) = (x^2 - x)(-2)\]\[ = -2(x^2 - x) = -2x^2 + 2x\]

Step 3: Integrate with respect to x

Now perform the outer integral with the result from Step 2: \[\int_{0}^{3} (-2x^2 + 2x) \, dx\]Calculate each term separately:\[\int (-2x^2) \, dx = -2\cdot \frac{x^3}{3}\Bigg|_{0}^{3} = -\frac{2}{3} [27 - 0] = -18 \]\[\int (2x) \, dx = 2\cdot \frac{x^2}{2}\Bigg|_{0}^{3} = [9 - 0] = 9\]Combine the results:\[-18 + 9 = -9\]

Step 4: Conclude the evaluation

The total value of the double integral over the specified region is \(-9\). This shows the net sum of the function \(x^2 y - 2xy\) over the described rectangle within the xy-plane.

Key Concepts

Region of Integration

When dealing with double integrals, understanding the region of integration is crucial. In this exercise, the limits for the inner and outer integrals define a rectangle in the xy-plane. The x limits are from 0 to 3, and the y limits are from -2 to 0. These form the boundaries of the region we integrate over.
Visualizing these boundaries, you get a rectangle with corners at

• (0, -2)
• (3, -2)
• (3, 0)
• (0, 0)

Starting from the lower-left corner, the rectangle expands rightward to x = 3 and upward to y = 0. This sketch helps clarify where the integration occurs.
Seeing this diagram allows you to verify the region is correctly determined before proceeding with calculations.

Inner Integral

The inner integral is the part where we first integrate with respect to y. In this exercise, the inner integral is

• \(\int_{-2}^{0} (x^2 y - 2xy) \, dy\)

For the inner integral, treat x as a constant since we integrate over y. As you integrate, be mindful of each part of the expression

• \(x^2 y\)
• \(- 2xy\)

Integrating these:

• \( x^2 \cdot \frac{y^2}{2} - 2x \cdot \frac{y^2}{2} = (x^2 - x)\left( \frac{y^2}{2} \right) \)

After integration, apply the limits from y = -2 to y = 0. Completing this step gives the expression

• -2(x^2 - x)

which is vital for the next phase of the double integration.

Outer Integral

Moving onto the outer integral, you'll use the result from the inner integration and integrate with respect to x. This makes the outer integral:

• \(\int_{0}^{3} (-2x^2 + 2x) \, dx\)

Handle each term separately:

• Integrating \(-2x^2\) yields \(-\frac{2}{3}x^3\)
• Integrating \(2x\) gives \(x^2\)

Evaluate each of these terms from x = 0 to x = 3:

• For \(-2x^2\): \(-\frac{2}{3}[x^3]\bigg|_{0}^{3} = -18\)
• For \(2x\): \(x^2\bigg|_{0}^{3} = 9\)

After evaluation, sum the results to get -9. This completion of the outer integral presents the solution to the double integration problem.

Evaluation of Integral

The evaluation of the integral is the culmination of integrating both the inner and outer integrals. This process combines the work from both steps to find the double integral's total value.
The evaluated value is

• -9

meaning if you integrate the function \(x^2 y - 2xy\) over the defined region, the result is -9.
This number represents the cumulative effect or "net sum" of the function across the region.
Reviewing this, double check each step of integration and substitution to ensure they align with the initial question and limits. Evaluating correctly ensures you can accurately describe the behavior of the function over its specified region.