Question

The region bounded below by the parabola $$y=x^{2}$$ and above $$b$$ . the line $$y=4$$ is to be partitioned into two subsections of equal area by cutting across it with the horizontal line $$y=c$$. (a) Sketch the region and draw a line $$y=c$$ across it that looks about right. In terms of $$c,$$ what are the coordinates of the points where the line and parabola intersect? Add them to your figure. (b) Find $$c$$ by integrating with respect to y. (This puts $$c$$ in the limits of integration. (c) Find $$c$$ by integrating with respect to $$x$$ . (This puts $$c$$ into the integrand as well.)

Short answer

The horizontal line that partitions the area into two equal parts is \(y=c=16/3\).

Step-by-step solution

Sketch the region and find intersection points

First, sketch the parabola \(y=x^{2}\) and the line \(y=4\). Draw a horizontal line \(y=c\) across the region. To find the coordinates of the intersection points of the horizontal \(y=c\) and the parabola \(y=x^{2}\), it should be set \(y=c\) equal to \(x^{2}\), this gives \(x = ±\sqrt{c}\).

Find \(c\) by integrating with respect to \(y\)

The area between \(y=c\) and \(y=4\) can be found by calculating the integral of the horizontal lines that form during integration with respect to \(y\). Integrate the expression \((4-c) * 2\sqrt{y}\) from \(c\) to \(4\) and set it equal to half the total area \(8-c^{2}/2\). Solving this equation yields \(c = 16/3\).

Find \(c\) by integrating with respect to \(x\)

Integrate from \(-\sqrt{c}\) to \(\sqrt{c}\) the expression \(c-x^{2}\), and from \(-2\) to \(2\) the expression \(4-x^{2}\). Setting these two integrals equal to each other and solving the resulting equation gives \(c = 16/3\).

Key concepts

Area under a curve

Understanding the concept of the area under a curve is essential when working with integrals in calculus. When we refer to the 'area under a curve', we generally mean the region between the graph of a function and the horizontal x-axis on a coordinate system. Calculating this area can be approached by partitioning the region into infinitesimally thin rectangles (or other shapes, depending on the method), then summing their areas—which is essentially what integration does.

In the context of our exercise, we're dealing with a parabola, which opens upwards, and the horizontal line at a height of 'b', which in this case is the line where y=4. By integrating properly with respect to the variable y, the total area that is enclosed between these two curves can be found. Moreover, we can even partition this area into two equal parts by introducing another horizontal line, y=c, and finding the specific value of c that accomplishes this division.

To visualize why this works, imagine the parabola as a mountain and the horizontal lines as flat terraces cut into the side. By finding the right level to place our terrace (our line y=c), we aim to have equal amounts of mountain above and below the terrace. This is an application of the area under the curve technique, and it uses definite integrals to find the precise value of c.

Definite Integrals

Definite integrals are all about finding the exact value of the area under a curve within a specific interval. When we calculate a definite integral, we are essentially summing up an infinite number of infinitesimal areas over a certain range, which together form the area under a function's curve on a graph.

In the given exercise, we're asked to partition an area beneath a parabola. The process involves two main steps of integration, first with respect to y, and then with respect to x, both of which involve setting up and evaluating definite integrals. Integrating with respect to y allows students to consider vertical slices of the area. It's somewhat like slicing a loaf of bread into thin, vertical pieces and then calculating the individual areas of these slices to find the whole.

When we integrate the expression \( (4-c) \cdot 2\sqrt{y} \) from c to 4, we're computing the vertical slices' individual areas and summing them up to get a half of the total area under the curve from y=c to y=4. In this exercise, finding the value of c that splits the area in two equal parts is tantamount to dividing a chocolate bar evenly, so that each person gets an equal share.

Symmetry in Parabolas

Parabolas are well-known for their symmetrical properties. The symmetry of a parabola is much like a mirror reflection; each side is a mirror image of the other about its line of symmetry. This line is also known as the axis of symmetry, which for the standard parabola \(y = x^2\) runs vertically along the y-axis.

In mathematical tasks, especially integration, symmetry can be a very powerful tool. It helps to simplify calculations and find solutions more efficiently. In our exercise, the use of symmetry is pivotal when we integrate with respect to x. Since the parabola is symmetrical about the y-axis, the areas to the left and right of the y-axis are equal, which leads to an easier time evaluating the integral required to find y=c.

When students integrate with respect to x from \( -\sqrt{c} \) to \( \sqrt{c} \), they are using the symmetrical property of the parabola to simplify their calculations. As the parabola's sections are reflective across the y-axis from -2 to 2, accounting for this balance in the problem provides a direct path to find the value for c that evenly splits the area. This allows for the integration to only be performed over half the parabola, doubling that result to get the entire area under the curve, which is an elegant application of symmetry in parabolas.