Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the indicated double integral over \(R\). $$ \iint_{R} x y^{3} d A ; R=\\{(x, y): 0 \leq x \leq 1,-1 \leq y \leq 1\\} $$

Step-by-Step Solution

Answer

The double integral over region \( R \) evaluates to 0.

Step 1: Understand the Region of Integration

The region \( R \) is defined as \( \{(x, y): 0 \leq x \leq 1, -1 \leq y \leq 1\} \), which is a rectangle on the xy-plane with x ranging from 0 to 1 and y ranging from -1 to 1.

Step 2: Set up the Double Integral

Based on the region \( R \), the double integral will be set up as the integral of \( xy^3 \) with respect to \( y \) first (from -1 to 1) and then \( x \) (from 0 to 1):\[\int_{0}^{1} \int_{-1}^{1} xy^3 \, dy \, dx\]

Step 3: Integrate with Respect to y

First, focus on the inner integral \( \int_{-1}^{1} xy^3 \, dy \). Treat \( x \) as a constant:\[ \int_{-1}^{1} y^3 \, dy = \left[ \frac{y^4}{4} \right]_{-1}^{1} \]Evaluate the integral:\[ \left( \frac{1^4}{4} \right) - \left( \frac{(-1)^4}{4} \right) = \frac{1}{4} - \frac{1}{4} = 0 \]

Step 4: Conclude Integration Over y

The result of the inner integral \( \int_{-1}^{1} y^3 \, dy \) is 0, which implies that the entire double integral will be zero because multiplication by zero in the next step for \( x \) integration will yield zero.

Step 5: Evaluate the Outer Integral

Since the inner integral evaluates to zero, no further integration needs to be done over \( x \) because any number multiplied by zero is zero. Thus, the problem simplifies to:\[ \int_{0}^{1} 0 \, dx = 0 \]

Step 6: Finalize the Solution

The double integral evaluates to:\[ \iint_{R} xy^3 \, dA = 0 \] because the y-component of the integration leads to a zero value due to symmetry of odd functions across bounds that are symmetrical around zero.

Key Concepts

Region of Integration

In double integrals, defining the region of integration is crucial. For this exercise, the region is given as \( R = \{(x, y): 0 \leq x \leq 1, -1 \leq y \leq 1\} \). This region represents a rectangular section on the xy-plane. The boundaries indicate the limits of the integral.

• For \( x \), the region starts at 0 and ends at 1.
• For \( y \), the region starts at -1 and ends at 1.

Understanding this region helps visualize where the integration takes place. It's like deciding the boundaries of a plot on a map before you plan what's inside. Each point \((x, y)\) within these boundaries contributes to the integral calculation. Recognizing these limits helps set up the double integral correctly.

xy-plane

The xy-plane is the two-dimensional space where our region of integration lies. It is a common reference space in mathematics that allows us to graphically represent the relationship between variables \( x \) and \( y \). Double integrals evaluated in this plane calculate volumes under surfaces, much like how a single integral finds areas under curves.

• The x-axis and y-axis form the coordinate grid where each point \((x, y)\) corresponds to a position in this plane.
• In this problem, the defined region \( R \) occupies a rectangular part of this plane.

Visualizing the problem on the xy-plane helps understand how functions behave across different regions. Seeing functions like \( xy^3 \) and their interactions over \( R \) aids in grasping the entirety of the integration problem.

Integration by Parts

Though not explicitly used in the given solution, Integration by Parts is a key technique often helpful in evaluating integrals. It is particularly useful when dealing with products of functions. The formula for integration by parts is derived from the product rule for differentiation and is expressed as:
\[ \int u \, dv = uv - \int v \, du \]where \( u \) and \( v \) are differentiable functions of \( x \). This technique transforms the integral of a product of two functions into simpler parts.

• Choose \( u \) to be a function that simplifies upon differentiation.
• Choose \( dv \) to be a function that, when integrated, remains manageable.

Understanding when and how to use integration by parts can greatly enhance efficiency in solving complex problems within calculus, making it a powerful tool for integrals that are products of multiple components.

Symmetry in Integration

Symmetry plays a crucial role in simplifying integrals. In this exercise, the function \( xy^3 \) exhibits symmetry concerning the y-axis due to the power of \( y \). This function involves an odd degree (3) of \( y \), making it symmetrical across the y-axis when considering the integral over \( y \) from \(-1\) to \(1\).

• Functions with odd powers are known to integrate to zero over symmetric limits centered at zero.
• In the exercise, \( \int_{-1}^{1} y^3 \, dy = 0 \) demonstrates this effectively.

Recognizing such symmetry can quickly lead to simplifications. It often means parts of the integral cancel out, significantly reducing computation. Comprehending symmetry in functions helps solve integrals more intuitively and efficiently.