Problem 56 - Thomas Calculus

Question

\(f(x, y)=x+2 y\) over the region \(R\) inside the circle \((x-2)^{2}+(y-3)^{2}=1\) using the partition \(x=1,3 / 2,2,5 / 2\) 3 and \(y=2,5 / 2,3,7 / 2,4\) with \(\left(x_{k}, y_{k}\right)\) the center (centroid) in the \(k\) th subrectangle (provided the subrectangle lies within \(R )\)

Step-by-Step Solution

Answer

Evaluate \( f(x, y) \) at each centroid within the circle, then sum these values.

Step 1: Understanding the Region

The region of interest, \( R \), is defined by the circle \((x-2)^2 + (y-3)^2 = 1\). This represents a circle centered at \((2, 3)\) with a radius of 1.

Step 2: Subdividing the Domain

We have been asked to partition the regions with \( x = 1, 3/2, 2, 5/2, 3 \) and \( y = 2, 5/2, 3, 7/2, 4 \). This forms rectangles or subregions within the whole domain.

Step 3: Checking Position

Determine if the center of each subrectangle lies within the region \( R \), the circle.

Step 4: Centroid Calculation

For each subrectangle, calculate the center \((x_k, y_k)\) using the midpoint formula of the interval on each axis. For example, the center of the subrectangle bounded by \(1 \leq x \leq \frac{3}{2}\) and \(2 \leq y \leq \frac{5}{2}\) is \( \left( \frac{1 + \frac{3}{2}}{2}, \frac{2 + \frac{5}{2}}{2} \right) = \left( \frac{5}{4}, \frac{9}{4} \right) \).

Step 5: Verifying Containment

Check whether each calculated centroid from the previous step falls inside the circle \((x-2)^2 + (y-3)^2 = 1\) by substituting \(x_k\) and \(y_k\) into the circle's equation.

Step 6: Evaluate Function

For each centroid that lies within the region, evaluate \(f(x_k, y_k) = x_k + 2y_k\).

Step 7: Conclusion

Sum up the values evaluated in Step 6 to estimate the total contribution from the subrectangles within \( R \).

Key Concepts

Centroid of a Region

Understanding the centroid of a region is crucial when working with double integration over a divided area. A centroid is the geometric center of an object, in this case, a subrectangle derived from a larger circle. Here, each subrectangle's centroid can be found using its intervals.For any subregion bounded by segments in two dimensions, the centroid's coordinates are calculated using the midpoint formula:

• The x-coordinate of the centroid is the average of the x-boundaries.
• The y-coordinate of the centroid is the average of the y-boundaries.

This gives us coordinates \( (x_k, y_k) \) for each subregion. Once we have these, we can verify whether each point lies inside the region we are analyzing, such as a circle, and if so, use it in further calculations. Calculating centroids precisely ensures accuracy in the whole process.

Circle Equations

Circle equations are key to understanding the boundaries of a region in a plane. The general form of a circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius. In this exercise, we are dealing with a circle centered at \( (2, 3) \) with a radius of 1.

• To check if a point lies within this circle, substitute the point's coordinates into the equation.
• If the equation holds true or is less than \( r^2, \) then the point lies inside or on the boundary.

This circle defines our region of interest, shaping the limits of our integration. Understanding this geometric boundary helps us filter which centroids are relevant for the function evaluation.

Function Evaluation

Function evaluation involves computing a given function’s value at specific points. In the context of this problem, the function being evaluated is \( f(x, y) = x + 2y \). Once the centroids are calculated and verified within the circle's boundaries, they are used as input to the function.For each valid centroid, we substitute its coordinates into the function to find its value:

• Compute \( f(x_k, y_k) = x_k + 2y_k \)
• This parameter represents an integral component contribution from that subregion to the overall result.

Each result is summed up to provide an approximation of the integral over the entire region. Evaluating functions accurately at these points is vital for a precise estimation.

Subregions

Subregions are smaller rectangles or squares that divide the overall integration region. These are particularly helpful when the main region, such as a circle, is complex, and subdividing it makes calculations manageable.Here, the exercise uses the partitions: \( x = 1, 3/2, 2, 5/2, 3 \) and \( y = 2, 5/2, 3, 7/2, 4 \). This creates multiple rectangular subregions.

• Each subregion’s boundary is given by one x-interval and one y-interval.
• Only those that are wholly or partially inside the circle are considered for further steps, including centroid and function evaluation.

Breaking down the region into these simpler parts allows for a clear and systematic way of evaluating an integral, ensuring no part of the larger region is ignored or double-counted.