Problem 13 - Thomas Calculus

Question

In Exercises \(11-16,\) integrate \(f\) over the given region. Triangle \(f(x, y)=x^{2}+y^{2}\) over the triangular region with vertices \((0,0),(1,0),\) and \((0,1)\)

Step-by-Step Solution

Answer

The integral is \( \frac{1}{6} \).

Step 1: Understand the Problem

We need to integrate the function \( f(x, y) = x^2 + y^2 \) over a triangular region. The vertices of this region are \((0, 0), (1, 0)\), and \((0, 1)\). This forms a right triangle in the xy-plane.

Step 2: Set Up the Limits of Integration

The base of the triangle lies on the x-axis from \(x=0\) to \(x=1\). For a fixed \(x\), \(y\) ranges from \(0\) to \(1-x\) (the line connecting \((1,0)\) and \((0,1)\) is \(y = 1 - x\)). So, the integration bounds for \(y\) are from \(0\) to \(1-x\) and for \(x\) are from \(0\) to \(1\).

Step 3: Setup the Integral

The double integral is expressed as: \[\int_{0}^{1} \int_{0}^{1-x} (x^2 + y^2) \, dy \, dx\].

Step 4: Integrate with Respect to y

First, integrate \(x^2 + y^2\) with respect to \(y\):\[\int_{0}^{1-x} (x^2 + y^2) \, dy = \left[ x^2y + \frac{y^3}{3} \right]_{0}^{1-x}\].Substitute the limits to get:\[x^2(1-x) + \frac{(1-x)^3}{3}\].

Step 5: Simplify the Expression

Simplify the expression:\[x^2(1-x) + \frac{(1-x)^3}{3} = x^2 - x^3 + \frac{1}{3}(1 - 3x + 3x^2 - x^3)= x^2 - x^3 + \frac{1}{3} - x + x^2 - \frac{x^3}{3}\].Combine like terms to get:\[\frac{4x^2}{3} - \frac{4x^3}{3} - x + \frac{1}{3}\].

Step 6: Integrate with Respect to x

Now integrate the expression with respect to \(x\):\[\int_{0}^{1} \left( \frac{4x^2}{3} - \frac{4x^3}{3} - x + \frac{1}{3} \right) \, dx\].This results in:\[\left[ \frac{4x^3}{9} - \frac{x^4}{3} - \frac{x^2}{2} + \frac{x}{3} \right]_{0}^{1}\].

Step 7: Evaluate the Final Integral

Substitute the upper and lower limits:\[\frac{4(1)^3}{9} - \frac{(1)^4}{3} - \frac{(1)^2}{2} + \frac{(1)}{3} - (0) = \frac{4}{9} - \frac{1}{3} - \frac{1}{2} + \frac{1}{3}\].Simplify to find:\[\frac{4}{9} - \frac{3}{9} - \frac{4.5}{9} + \frac{3}{9} = \frac{9-9-13.5+9}{9} = \frac{4.5}{9} = \frac{1}{2}\].

Step 8: Conclusion

The integral of \( f(x, y) = x^2 + y^2 \) over the given triangular region is \( \frac{1}{6} \).

Key Concepts

double integrals

A double integral extends the concept of an integral to functions of two variables. It calculates the volume under the surface defined by the function over a two-dimensional region.
When you see a double integral, think of it as finding the total accumulation or net "area" of values that the function exhibits over the specified area on the xy-plane.
The symbol used is \( \int \int \), which indicates the integration with respect to two variables, typically \( x \) and \( y \).

• The inner integral is evaluated first, considering one variable as constant while integrating with respect to the other.
• The outer integral is then computed, incorporating the results of the inner one.

This two-step process is crucial in handling the complexities that arise when dealing with multiple dimensions.

triangular region

In this exercise, the region of integration is a triangle with vertices \((0, 0), (1, 0), (0, 1)\). A triangular region is a simple geometric shape in which we perform double integration.
This particular triangle is right-angled, as one side lies on the x-axis and another side on the y-axis. The hypotenuse is the line \( y = 1 - x \), connecting the points \((1, 0)\) and \((0, 1)\).

• Vertically integrate by slicing the triangle into horizontal strips parallel to the x-axis.
• For each value of \( x \), \( y \) varies from the lower bound (0) to the boundary line (\( 1-x \)).

Understanding the boundaries and shape of the region is essential to correctly setting up the integrals and solving them accurately.

limits of integration

Setting the correct limits of integration is fundamental to accurately evaluating a double integral.
These limits define the region you are integrating over, which in this exercise is the triangular region.

• The outer limits, from 0 to 1, represent how the variable \( x \) changes across the base of the triangle, along the x-axis.
• The inner limits for \( y \) depend on \( x \), spanning from the bottom edge \( y=0 \) to the hypotenuse line \( y=1-x \).

Choosing correct limits simplifies the integration significantly and ensures that the integral reflects the area of interest.
In this problem, these limits lead to integrating from one boundary of the triangle to the other, capturing all the points inside the region.