Problem 16 - Calculus for Biology and Medicine

Question

Find the areas of the regions bounded by the lines and curves by expressing \(x\) as a function of \(y\) and integrating with respect to \(y .\) \(x=(y-1)^{2}-1, x=(y-1)^{2}+1\) from \(y=0\) to \(y=2\)

Step-by-Step Solution

Answer

The area of the region is 4 square units.

Step 1: Understanding the Curves

First, let's rewrite the equations given: \( x = (y-1)^2 - 1 \) and \( x = (y-1)^2 + 1 \). These represent two parabolas that open to the right, both based on the expression \( x = (y-1)^2 \), shifted vertically.

Step 2: Determine Intersection Points

The lines \( x = (y-1)^2 - 1 \) and \( x = (y-1)^2 + 1 \) intersect when their expressions for \( x \) are equal. Set \( (y-1)^2 - 1 = (y-1)^2 + 1 \). Simplifying this gives \( -1 = 1 \), which is a contradiction. Therefore, the curves do not intersect except at vertical lines.

Step 3: Identify the Region Bounded by the Curves

Between \( y = 0 \) and \( y = 2 \), the region we are interested in is between the curves \( x = (y-1)^2 + 1 \) (the right line) and \( x = (y-1)^2 - 1 \) (the left line).

Step 4: Set up the Integral

We integrate with respect to \( y \) from 0 to 2 the difference between the right and left expressions: \( \int_{0}^{2} \left( ((y-1)^2 + 1) - ((y-1)^2 - 1) \right) \, dy \).

Step 5: Simplify the Integrand

The integrand simplifies as follows: \( ((y-1)^2 + 1) - ((y-1)^2 - 1) = 2 \). Hence, the integral becomes \( \int_{0}^{2} 2 \, dy \).

Step 6: Evaluate the Integral

Evaluating \( \int_{0}^{2} 2 \, dy \) gives \( 2y \) from 0 to 2. Substitute the limits: \( 2(2) - 2(0) = 4 \).

Step 7: Conclude the Solution

The area of the region bounded by the curves from \( y = 0 \) to \( y = 2 \) is 4 square units.

Key Concepts

Area Between Curves

When finding the area between curves, the key is to identify the region bounded by the relevant curves or lines. These areas can be calculated by integrating the difference between the functions of the curves. For the exercise in question, we're given two curves:

• \( x = (y-1)^2 - 1 \)
• \( x = (y-1)^2 + 1 \)

These represent two parabolas that we need to compare for the range of \( y \) from 0 to 2. To find the area between these curves, we set up an integral with respect to \( y \) of the difference between the rightmost and the leftmost functions. This integral is taken over the interval where these curves are defined and do not intersect each other vertically.

• Right curve: \( x = (y-1)^2 + 1 \)
• Left curve: \( x = (y-1)^2 - 1 \)

The area is calculated by integrating \( 2 \) over the \( y \)-range from 0 to 2, giving an area of 4 square units. Always remember, the key is the difference between the outer and inner curves, as this gives the height of the rectangular strip used in integration.

Parabolas

Parabolas are a type of curve that can be expressed through quadratic equations. In this exercise, parabolas open to the right because they take the form \( x = (y-1)^2 - c \), where \( c \) is a constant. The generic form \( x = a(y-b)^2 + c \) represents a parabola centered around the vertex \((c, b)\), with "a" controlling the width and direction of the opening. In our specific example:

• The parabolas are centered shifted by 1 unit because of the term \((y-1)^2\).
• The shifting in the \( x \)-direction (horizontally) is determined by the constants \(-1\) and \(+1\), added to the square term.

These parabolas are symmetric about the vertex point on the line \( y = 1 \), making it easier to see how the parabolas relate and overlap over the specified range of \( y \). Understanding parabolas is critical for visualizing and setting up integrals, particularly when using them to calculate areas between curves.

Integrate with Respect to y

Integrating with respect to \( y \) might seem unusual at first, especially if you're accustomed to integrating with respect to \( x \). However, it can often simplify problems involving vertical strips and x-boundaries. Here, the integration is performed over the \( y \)-interval \( [0, 2] \). This choice is necessary when given functions where \( x \) is expressed explicitly in terms of \( y \), rather than \( y \) in terms of \( x \).

To set up the integral for the area, compute:\[\int_{0}^{2} [(y-1)^2 + 1] - [(y-1)^2 - 1] \, dy \]This simplifies to:\[\int_{0}^{2} 2 \, dy \]This approach highlights the upper and lower bounds on \( y \) and focuses directly on how the functions change with \( y \). Because no intersections occur, the limits on \( y \) remain straightforward. Trusting in the mechanics of integration truly brings out the elegant simplicity in calculus when dealing with functions that depend on a single variable, reminding us of calculus' power and flexibility.