Question

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{1}^{3} \int_{y}^{3}\left(x^{2} y-x y^{2}\right) d x d y $$

Short answer

The value of the integral is 4.1667; in reverse, \( \int_{1}^{3} \int_{1}^{x} (x^2 y - xy^2) \, dy \, dx \)."

Step-by-step solution

Evaluate the Inner Integral

Let's evaluate the inner integral with respect to \(x\). Given that the inner integral is \( \int_{y}^{3} (x^2 y - x y^2) \, dx \), we need to integrate each term separately:1. The first term \( x^2 y \) integrates to \( \frac{1}{3} x^3 y \).2. The second term \( -x y^2 \) integrates to \( -\frac{1}{2} x^2 y^2 \).Substitute the limits of integration:\[ \int_{y}^{3} (x^2 y - x y^2) \, dx = \left[ \frac{1}{3} x^3 y - \frac{1}{2} x^2 y^2 \right]_y^3. \]

Substitute Upper and Lower Limits for x

Evaluate the expression from Step 1 by substituting the upper limit \(x = 3\) and lower limit \(x = y\):- Substituting \(x = 3\): \[ \frac{1}{3} (3)^3 y - \frac{1}{2} (3)^2 y^2 = 9y - \frac{9}{2} y^2 \]- Substituting \(x = y\): \[ \frac{1}{3} y^3 y - \frac{1}{2} y^2 y^2 = \frac{1}{3} y^4 - \frac{1}{2} y^4 = -\frac{1}{6} y^4 \]Subtract the second result from the first:\[ (9y - \frac{9}{2} y^2) - (-\frac{1}{6} y^4) = 9y - \frac{9}{2} y^2 + \frac{1}{6} y^4. \]

Evaluate the Outer Integral

Now integrate the expression from Step 2 with respect to \(y\):\[ \int_{1}^{3} \left( 9y - \frac{9}{2} y^2 + \frac{1}{6} y^4 \right) \, dy. \]Integrate term by term:1. \( 9y \) integrates to \( \frac{9}{2} y^2 \).2. \( -\frac{9}{2} y^2 \) integrates to \( -\frac{9}{6} y^3 = -\frac{3}{2} y^3 \).3. \( \frac{1}{6} y^4 \) integrates to \( \frac{1}{30} y^5 \).Substitute the limits of integration:\[ \left[ \frac{9}{2} y^2 - \frac{3}{2} y^3 + \frac{1}{30} y^5 \right]_1^3 \]

Substitute the Limits for y

Evaluate the expression derived in Step 3 at the limits \(y = 3\) and \(y = 1\):- At \(y = 3\): \[ \frac{9}{2} (3)^2 - \frac{3}{2} (3)^3 + \frac{1}{30} (3)^5 = 40.5 - 40.5 + 7.2 = 7.2 \]- At \(y = 1\): \[ \frac{9}{2} (1)^2 - \frac{3}{2} (1)^3 + \frac{1}{30} (1)^5 = 4.5 - 1.5 + 0.0333 = 3.0333 \]Subtract the values:\[ 7.2 - 3.0333 = 4.1667 \]

Rewrite with Reverse Order of Integration

To change the order of integration, consider the region of integration in the \(xy\)-plane. The original bounds are \(1 \leq y \leq 3\) and \(y \leq x \leq 3\). The new order requires us to evaluate \(x\) first:- The limits for \(y\) are \(1 \leq y \leq x\). - The limits for \(x\) are \(1 \leq x \leq 3\).The rewritten integral is:\[ \int_{1}^{3} \int_{1}^{x} (x^2 y - xy^2) \, dy \, dx \]

Final Answer

The value of the iterated integral is approximately 4.1667. The rewritten integral in the opposite order is \( \int_{1}^{3} \int_{1}^{x} (x^2 y - xy^2) \, dy \, dx \).

Key concepts

Integration by Parts

Integration by parts is a technique used to integrate products of functions. In this exercise, we aren't explicitly using integration by parts, but understanding the formula is crucial for more complex integrals you might encounter. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

You start by identifying parts of the integral to serve as \(u\) and \(dv\), then differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\). After substituting these into the formula, you can solve an integral easier than the original. While this exercise deals with iterated integrals directly through integration by terms, familiarity with integration by parts can deepen your understanding and may come in handy for solving related problems.

Order of Integration

The order of integration refers to the sequence in which you integrate multiple variables. This is crucial in iterated integrals, where you deal with complex functions of two or more variables.

• In the exercise, the order of integration was initially \(\int_{1}^{3} \int_{y}^{3} (x^2 y - x y^2) \, dx \, dy \), meaning you integrate with respect to \(x\) first, and then \(y\).
• Later, this order is reversed to \( \int_{1}^{3} \int_{1}^{x} (x^2 y - xy^2) \, dy \, dx \). Here, you begin integrating with respect to \(y\) and then with respect to \(x\).

Changing the order of integration can make the integral easier to solve, especially if one order simplifies the calculation or better accommodates the function's limits. Understanding this concept allows flexibility and efficiency when evaluating double integrals.

Limits of Integration

Limits of integration define the bounds within which you integrate. These limits depend on the region in the plane where your function is defined.

• In the original problem, the limits for \(y\) were from 1 to 3, and the limits for \(x\) were dynamic, dependent on \(y\), ranging between \(y\) and 3.
• When we reversed the order of integration, the limits also changed. Now, for \(y\), limits ranged from 1 to \(x\), and for \(x\), it ranged from 1 to 3.

Accurately identifying and using the correct limits of integration are essential for solving integrals because they define the scope of evaluation in the given region. In iterated integrals, especially, adjusting these limits correctly when changing the order of integration ensures precision in calculating areas or volumes.

Double Integrals

Double integrals allow us to compute the volume under a surface defined over a region in the plane. In mathematical terms, if you have a function \(f(x, y)\) defined on a region \(R\), then the double integral \(\int \int_R f(x, y) \, dA\) finds the accumulated sum of values \(f(x, y)\) over that region.

• In this exercise, the function \((x^2 y - xy^2)\) is evaluated over a specific rectangular region with variable limits.
• Each integration corresponds to summing along one dimension of the region: first along \(x\), then \(y\), or vice versa.

Understanding how to set up and evaluate double integrals is critical, as it expands your toolkit for solving problems involving areas and volumes in multivariable calculus. This provides a powerful method for integrating over regions bounded by more complex functions or geometric shapes.