Area Of Region R Calculated With Horizontal Rectangles
In the realm of calculus, determining the area of a region bounded by curves is a fundamental concept. While vertical rectangles are a common approach, exploring horizontal rectangles offers a unique perspective and, in some cases, a more elegant solution. This article delves into the process of calculating the area of a region R bounded by the graphs of the functions f(x) = 3x², g(x) = 4 - x, and the line y = 0 using horizontal rectangles. We'll break down the steps, highlighting the key considerations and providing a comprehensive understanding of this technique.
Understanding the Region R
Before diving into the calculations, it's crucial to visualize the region R. The function f(x) = 3x² represents a parabola opening upwards, with its vertex at the origin (0, 0). The function g(x) = 4 - x represents a straight line with a slope of -1 and a y-intercept of 4. The line y = 0 represents the x-axis. To fully grasp the shape of R, we need to find the points of intersection between these curves. First, let's find the intersection points between the parabola and the line. We set f(x) = g(x), which gives us 3x² = 4 - x. Rearranging this equation, we get 3x² + x - 4 = 0. This quadratic equation can be factored as (3x + 4)(x - 1) = 0, yielding solutions x = -4/3 and x = 1. Since we are considering the region bounded by y = 0, we are interested in the intersection point where x is positive. Thus, the relevant intersection point is at x = 1. At this point, f(1) = 3(1)² = 3 and g(1) = 4 - 1 = 3, so the curves intersect at the point (1, 3). Next, we find where each curve intersects the x-axis (y = 0). For the parabola, 3x² = 0 implies x = 0. For the line, 4 - x = 0 implies x = 4. Therefore, the parabola intersects the x-axis at (0, 0), and the line intersects the x-axis at (4, 0). With these intersection points in mind, we can sketch the region R. It's a region bounded by the parabola on the left, the line on the right, and the x-axis below. The region extends from x = 0 to x = 1 along the parabola and from x = 1 to x = 4 along the line. Now that we have a clear picture of the region R, we can proceed to calculate its area using horizontal rectangles.
Setting Up the Integral with Horizontal Rectangles
The essence of using horizontal rectangles lies in integrating with respect to y rather than x. This means we need to express our functions in terms of y. For f(x) = 3x², we solve for x in terms of y. Dividing by 3, we get x² = y/3. Taking the square root of both sides, we have x = ±√(y/3). Since we are considering the right side of the parabola (within our region R), we take the positive root, so x = √(y/3). This represents the parabolic boundary of our region as a function of y. For g(x) = 4 - x, we solve for x in terms of y. Adding x and subtracting y from both sides, we get x = 4 - y. This represents the linear boundary of our region as a function of y. When using horizontal rectangles, the width of each rectangle is dy, and the length is the difference between the x-values of the right boundary and the left boundary. In our region R, the right boundary is defined by the line x = 4 - y, and the left boundary is defined by the parabola x = √(y/3). Therefore, the length of a horizontal rectangle at a given y is (4 - y) - √(y/3). The region R extends from y = 0 (the x-axis) up to the point of intersection (1, 3), so our integration limits will be from y = 0 to y = 3. The area of the region R can then be expressed as the definite integral of the length of the horizontal rectangles with respect to y from 0 to 3. This gives us the integral:
Area = ∫[0 to 3] ((4 - y) - √(y/3)) dy
This integral represents the sum of the areas of infinitesimally thin horizontal rectangles within the region R. By evaluating this integral, we will find the exact area of the region.
Evaluating the Integral
Now, let's evaluate the integral we set up in the previous section:
Area = ∫[0 to 3] ((4 - y) - √(y/3)) dy
To solve this, we can break it into separate integrals:
Area = ∫[0 to 3] (4 - y) dy - ∫[0 to 3] √(y/3) dy
First, we evaluate the integral of (4 - y):
∫[0 to 3] (4 - y) dy = [4y - (1/2)y²] evaluated from 0 to 3
Plugging in the limits of integration, we get:
[4(3) - (1/2)(3)²] - [4(0) - (1/2)(0)²] = 12 - 4.5 = 7.5
Next, we evaluate the integral of √(y/3). We can rewrite this as (1/√3)√y:
∫[0 to 3] √(y/3) dy = (1/√3) ∫[0 to 3] y^(1/2) dy
Now, we find the antiderivative of y^(1/2), which is (2/3)y^(3/2). So, we have:
(1/√3) ∫[0 to 3] y^(1/2) dy = (1/√3) [(2/3)y^(3/2)] evaluated from 0 to 3
Plugging in the limits of integration, we get:
(1/√3) [(2/3)(3)^(3/2) - (2/3)(0)^(3/2)] = (1/√3) [(2/3)(3√3)] = 2
Now, we subtract the second integral from the first integral to find the area:
Area = 7.5 - 2 = 5.5
Therefore, the area of the region R is 5.5 square units.
Benefits of Using Horizontal Rectangles
In this particular problem, using horizontal rectangles proved to be a straightforward approach. While vertical rectangles could also be used, they would require splitting the region into two subregions because the top bounding function changes at x = 1. With horizontal rectangles, we were able to express the area as a single integral, simplifying the calculation. This highlights a key advantage of choosing the right method for area calculation. Horizontal rectangles are particularly useful when the functions are easily expressed in terms of y and when the region is more easily defined with respect to vertical lines (constant y-values) than horizontal lines (constant x-values). Understanding when to employ horizontal rectangles can significantly streamline the process of finding areas bounded by curves. By recognizing the geometric characteristics of the region and the functions involved, we can make an informed decision about the most efficient integration method.
Conclusion
Calculating the area of a region bounded by curves is a fundamental skill in calculus, and the choice of using vertical or horizontal rectangles can greatly impact the complexity of the solution. In this article, we successfully calculated the area of the region R bounded by f(x) = 3x², g(x) = 4 - x, and y = 0 using horizontal rectangles. We first visualized the region and found the points of intersection, then set up the integral with respect to y, and finally evaluated the integral to find the area. The result, 5.5 square units, demonstrates the effectiveness of this technique. By understanding the benefits of horizontal rectangles and applying them strategically, we can solve area problems more efficiently and gain a deeper appreciation for the power of integral calculus.
