Problem 13 - Advanced Engineering Mathematics

Question

Evaluate the double integral over the region \(R\) that is bounded by the graphs of the given equations. Choose the most convenient order of integration. $$ \iint_{R} x^{3} y^{2} d A ; y=x, y=0, x=1 $$

Step-by-Step Solution

Answer

The value of the double integral is \( \frac{1}{21} \).

Step 1: Understand the Region of Integration

The region \( R \) is defined by the curves \( y = x \), \( y = 0 \), and \( x = 1 \). This forms a triangular region in the xy-plane, with the vertices at \((0,0)\), \((1,0)\), and \((1,1)\).

Step 2: Decide the Order of Integration

Looking at the region of integration, it is easier to integrate first with respect to \( y \), followed by \( x \). This way, \( y \) varies from \( y=0 \) to \( y=x \), while \( x \) varies from \( x=0 \) to \( x=1 \).

Step 3: Set Up the Integral

With the chosen order of integration, the double integral can be expressed as: \[ \int_{0}^{1} \int_{0}^{x} x^{3} y^{2} \, dy \, dx. \]

Step 4: Integrate with Respect to y

First, integrate \( x^{3}y^{2} \) with respect to \( y \): \[ \int_{0}^{x} x^{3} y^{2} \, dy = x^3 \left[ \frac{y^3}{3} \right]_{0}^{x} = x^3 \left( \frac{x^3}{3} \right) = \frac{x^6}{3}. \]

Step 5: Integrate with Respect to x

Now integrate \( \frac{x^6}{3} \) with respect to \( x \): \[ \int_{0}^{1} \frac{x^6}{3} \, dx = \frac{1}{3} \int_{0}^{1} x^6 \, dx = \frac{1}{3} \left[ \frac{x^7}{7} \right]_{0}^{1} = \frac{1}{3} \left( \frac{1}{7} \right) = \frac{1}{21}. \]

Step 6: Conclusion

The value of the double integral \( \iint_{R} x^3 y^2 \, dA \) over the specified region is \( \frac{1}{21} \).

Key Concepts

Region of Integration

When dealing with double integrals, understanding the region of integration is paramount. In this exercise, the region of integration, denoted as \( R \), is enclosed by the lines \( y = x \), \( y = 0 \), and \( x = 1 \). This effectively forms a right triangle in the xy-plane.
The vertices of this triangular region are at points \((0,0)\), \((1,0)\), and \((1,1)\).

• The line \( y = 0 \) acts as the horizontal boundary along the x-axis from \( x=0 \) to \( x=1 \).
• The line \( y = x \) ascends diagonally from the origin to the point \((1,1)\).
• Finally, \( x = 1 \) is a vertical line that closes the triangle by connecting \((1,0)\) to \((1,1)\).

Visualizing this triangular region is essential for setting up the correct limits of integration, ensuring that the integration covers the entire defined section.

Order of Integration

The order of integration matters significantly when setting up a double integral. It determines how we slice the region \( R \) as we integrate. In this example, choosing to integrate first with respect to \( y \) and then \( x \) is strategic.

• First, consider fully slicing vertically. When integrating with respect to \( y \), \( y \) begins from the boundary \( y=0 \) and extends to the line \( y=x \).
• For \( x \), it consistently ranges from the minimum boundary of 0 to the maximum value of 1 throughout the integration.

This approach, often denoted as \( dydx \), is chosen because it simplifies integration by aligning with the natural boundaries present in the region. Understanding how these orders can simplify calculations is vital when approaching similar problems.

Integration Techniques

Integrating functions through double integrals often requires specific techniques for efficiency. In this problem, integrating step-by-step starts with the inner integral.
**Inner Integration**

• We begin integrating \( x^3 y^2 \) first with respect to \( y \). Treat \( x^3 \) as a constant because it does not depend on \( y \). This makes the integration process straightforward.
• Integrating \( y^2 \) results in \( \frac{y^3}{3} \). The substitution of limits of \( y \) from 0 to \( x \) yields \( \frac{x^6}{3} \), through the calculation \( x^3 \cdot \frac{x^3}{3} \).

**Outer Integration**

• We now integrate \( \frac{x^6}{3} \) with respect to \( x \).
• Repeat the process: find the antiderivative and evaluate from the given limits 0 to 1. Thus integrating \( x^6 \) gives \( \frac{x^7}{7} \), hence the outcome \( \frac{1}{21} \) for the entire double integral.

Mastering these steps is key to handling any double integral, especially when order and regional boundaries are distinct as seen here.

Changing Variables in Integrals

Sometimes, switching variables in a double integral can simplify the process or make complex regions easier to work with. While this wasn't necessary for our triangular region here, it's a useful technique worth mentioning.
**Why Change Variables?**

• Changing variables can transform a complex region into a simpler one, making integration more straightforward.
• This technique often involves variables substitutions or transformations that align better with symmetry or specific boundaries.

**How to Change Variables**Changing variables usually entails three main steps:

• Identify suitable new variables, say \( u \) and \( v \), linked to \( x \) and \( y \) with functional relationships, like \( u = x+y \), \( v = x-y \).
• Translate the limits and boundaries of \( x \) and \( y \) to fit \( u \) and \( v \).
• Calculate the Jacobian determinant to adjust the differential area \( dA \) accordingly in the new variable space.

While the change in variables wasn't applied in this exercise, familiarizing with this concept enhances an understanding of more complex integrals.