Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. \(\int_{1}^{2} \int_{0}^{3}\left(x y+y^{2}\right) d x d y\)

Step-by-Step Solution

Answer

The value of the integral is \( \frac{55}{4} \).

Step 1: Understand the Integral

We are given a double integral \( \int_{1}^{2} \int_{0}^{3}\left(x y+y^{2}\right) d x d y \). This represents the volume under the surface \( z = xy + y^2 \) between \( x = 0 \) to \( x = 3 \) and \( y = 1 \) to \( y = 2 \). We will first integrate with respect to \( x \).

Step 2: Integrate with Respect to x

Evaluate the inner integral \( \int_{0}^{3} (xy + y^2) \, dx \). Since \( y \) is treated as a constant with respect to \( x \), the integration is straightforward. \[ \int_{0}^{3} (x y + y^2) \, dx = \int_{0}^{3} x y \, dx + \int_{0}^{3} y^2 \, dx \]For \( \int_{0}^{3} x y \, dx = y \left[ \frac{x^2}{2} \right]_0^3 = y \left( \frac{9}{2} - 0 \right) = \frac{9y}{2} \).For \( \int_{0}^{3} y^2 \, dx = y^2 \left[ x \right]_0^3 = y^2 (3 - 0) = 3y^2 \).Thus, \[ \int_{0}^{3} (xy + y^2) \, dx = \frac{9y}{2} + 3y^2 \].

Step 3: Integrate with Respect to y

Now, substitute the result from Step 2 into the outer integral:\[ \int_{1}^{2} \left( \frac{9y}{2} + 3y^2 \right) \, dy \].Evaluate each part separately:For \( \int_{1}^{2} \frac{9y}{2} \, dy = \frac{9}{2} \left[ \frac{y^2}{2} \right]_1^2 = \frac{9}{2} \left( \frac{4 - 1}{2} \right) = \frac{9}{2} \times \frac{3}{2} = \frac{27}{4} \).For \( \int_{1}^{2} 3y^2 \, dy = 3 \left[ \frac{y^3}{3} \right]_1^2 = \left[ 8 - 1 \right] = 7 \).Thus, \[ \int_{1}^{2} \left( \frac{9y}{2} + 3y^2 \right) \, dy = \frac{27}{4} + 7 \].

Step 4: Simplify the Final Result

Combine the results from Step 3 to find the integral. Convert 7 to a fraction with a denominator of 4 for easy addition:\[ 7 = \frac{28}{4} \]Thus, the complete integral is:\[ \frac{27}{4} + \frac{28}{4} = \frac{55}{4} \].

Key Concepts

Iterated Integrals

Iterated integrals are a specific way of evaluating double integrals where one integration is performed within the limits of another. This technique breaks down a complex problem into manageable parts by repeatedly applying single-variable integration.
In the given exercise, we were asked to evaluate the iterated integral \( \int_{1}^{2} \int_{0}^{3}(xy + y^2) \, dx \, dy \). First, we focus on the innermost integral, integrating with respect to \( x \), and then, substitute this result into the outer integral with respect to \( y \).
This method simplifies solving planar regions and helps us build towards finding the volume under a surface by integrating section by section.

Integration with Respect to Variables

In iterated integrals, integration often occurs with respect to different variables at every step. For the exercise, we started by integrating with respect to the variable \( x \).
This means treating other variables, in this case, \( y \), as constants during the integration. Solving \( \int_{0}^{3} (xy + y^2) \, dx \) involved integrating the terms \( xy \) and \( y^2 \) separately, giving results based on treating \( y \) as a constant.
In the next step, the focus shifts, and integration is performed with respect to \( y \), involving handling specific variable powers, sums, or products. It shows the systematic approach necessary for handling multi-variable calculus problems.

Calculus Problem Solving

Solving calculus problems with double integrals is a vital skill in understanding volumes and areas in higher dimensions.
The process involves understanding the function describing the surface and the region over which integration must occur. This requires selecting appropriate limits and order of integration to break down complex shapes.
Breaking complex problems into single-variable integrals, evaluating step-by-step, and re-composing results is a key strategy in calculus that simplifies more abstract mathematical challenges.

Volume Under a Surface

One application of double integrals is finding the volume under a surface defined by a function \( z = f(x, y) \). This involves integrating over the defined region in the xy-plane.
In this exercise, the function \( z = xy + y^2 \) describes the surface, and the limits dictate the region. By integrating from \( y = 1 \) to \( y = 2 \) and \( x = 0 \) to \( x = 3 \), the resulting value represents the volume beneath this surface within given bounds.
This technique extends beyond simple shapes, allowing for calculation of volumes under more complex surfaces, which is useful in fields like physics or engineering.