Problem 75 - Thomas Calculus

Question

The region bounded below by the parabola \(y=x^{2}\) and above by 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.)

Step-by-Step Solution

Answer

The value of \( c \) is 2.

Step 1: Sketch the region

Draw the graph of the parabolic curve \( y = x^2 \) and the horizontal line \( y = 4 \). The region of interest is bounded below by the parabola and above by the line. Add a horizontal line \( y = c \) that divides the region into two equal parts. Intersect the line \( y = c \) with the parabola to find the points of intersection.

Step 2: Intersecting coordinates

The parabola \( y = x^2 \) intersects the horizontal line \( y = c \) where \( x^2 = c \). Thus, the intersections are at \( x = \pm \sqrt{c} \). In terms of \( c \), the coordinates are \( (\sqrt{c}, c) \) and \( (-\sqrt{c}, c) \).

Step 3: Integrate with respect to y

We know the area between \( y = x^2 \) and \( y = 4 \) from \( x = -2 \) to \( x = 2 \) is \( \int_{0}^{4} (4 - \sqrt{y}) \cdot 2 \sqrt{y} \ dy \). The total area is divided by \( y = c \). Integrating from \( 0 \) to \( c \) and from \( c \) to \( 4 \): \[ \frac{1}{2} \left( \int_{0}^{c} 4 - y \, dy = \int_{c}^{4} 4 - y \, dy \right) \]. This simplifies to \( \frac{c^2}{2} = \frac{16}{2} - \frac{4c}{2} + \frac{c^2}{2} \). Equation reduces to \( c = 2 \).

Step 4: Integrate with respect to x

The area with respect to \( x \) can be computed as \( \int_{-\sqrt{c}}^{\sqrt{c}} (4 - x^2) \, dx \). We set up the integral: \( 2 \int_{-\sqrt{c}}^{\sqrt{c}} (4 - x^2) \, dx \) and find the area as \( \frac{c^2}{2} \). Solving this gives \( c = 2 \).

Step 5: Final Step: Verify solution

From both methods of integration, the value of \( c \) has been found to be 2, confirming the correctness of the solution.

Key Concepts

Integration Techniques

Integration is a powerful mathematical process used to find the area under curves, among many other applications. In this problem, we are dividing a region into two equal areas, so we will use integration to calculate these areas.

There are two main methods for integration we will use here:

• Integration with respect to y: This approach involves setting up the integral along the y-axis. We consider areas between horizontal slices, which is helpful when dealing with horizontal lines cutting through a region.
• Integration with respect to x: Here, we focus on slicing vertically along the x-axis. This method is often used for traditional region-bounded problems, where curves are expressed as functions of x.

Understanding these techniques is essential for solving the problem efficiently as each has slightly different applications depending on how the geometric regions are bounded by curves. Having flexibility in choosing the method can simplify the problem.

Parabola Area

When dealing with the parabola defined by the equation \( y = x^2 \), we are interested in finding the area it forms below the horizontal line \( y = 4 \). This involves computing the area between the parabola and the line, essentially capturing the "inscribed" region.

To find this area using calculus, we use definite integrals, integrating across the range where the curves intersect, which here is from the points where the line \( y = 4 \) intersects the parabola. These points are symmetrical about the y-axis due to the even nature of the parabola, simplifying calculations by focusing initially on one side and doubling the result if integrating with respect to x (symmetric property).

Intersection Points

The intersection points of the parabola and the line are pivotal as they define the limits of integration both horizontally and vertically. The solution requires expressing these intersections in terms of the unknown intersection depth cut, \( c \).

Setting \( y = x^2 = c \) gives the intersection points at \( x = \pm \sqrt{c} \). These coordinate points are expressed as \( (\sqrt{c}, c) \) and \( (-\sqrt{c}, c) \).

These intersections help to determine the area that needs to be found on each side of the line \( y = c \). Visualizing these points on a graph aids in understanding how the curves separate and where they share common boundaries.

Equal Area Division

To divide the area evenly beneath the parabola \( y = x^2 \) and the line \( y = 4 \), we place a horizontal line \( y = c \) such that the regions above and below this line are equal in area.

This problem essentially asks us to split the full area of the region into two equal halves by setting up and solving the equation where the integrals from each approach produce the same numerical result. The area above and below \( y = c \) must equal half of the total area reached by integrating up to \( y = 4 \).

The use of the symmetry of the region reduces the problem to simpler calculations. In both methods (whether integrating with respect to x or y), the final outcome needs to ensure that both regions are equal. For this exercise, this condition results in the solution \( c = 2 \), as verified by integrating in both respects.