Problem 9 - Essential Calculus

Question

In Exercises 9 and 10 , find the area of the region by integrating (a) with respect to \(x\) and (b) with respect to \(y\). $$ \begin{array}{l} x=4-y^{2} \\ x=y-2 \end{array} $$

Step-by-Step Solution

Answer

The area of the region under the curves when integrating with respect to \(x\) is \(16/3\) and when integrating with respect to \(y\) is \(25/3\).

Step 1: Find points of intersection

Firstly, set the two equations equal to each other to find the points of intersection. Solve the equation \(4 - y^{2} = y - 2\) which can be rewritten as \(y^{2} + y - 6 = 0\). This is a quadratic equation and can be solved by factoring to get \(y - 2\)(\(y + 3\) = 0. So, \(y = 2\) and \(y = -3\) are where the curves intersect.

Step 2: Integrate with respect to x

For integrating with respect to \(x\), solve the equation \(x = 4 - y^{2}\) for \(y\). You will get two solutions \(y = \sqrt{4 - x}\) and \(y = -\sqrt{4 - x}\). Find the definite integral of the difference of these equations on the interval \(-2 \leq x \leq 2\). Thus \(\int_{-2}^{2} [\sqrt{4 - x} - (-\sqrt{4 - x})] dx = 16/3\) is the area under the curve when integrated with respect to \(x\).

Step 3: Integrate with respect to y

For integrating with respect to \(y\), you need the lower limit as lower function and upper limit as upper function. So take \(y - 2\) as the lower limit and \(4 - y^{2}\) as the upper limit. Find the definite integral from \(-3 \leq y \leq 2\), thus \(\int_{-3}^{2} [(4 - y^{2}) - (y - 2)] dy = 25/3\) which is the area under the curve when integrated with respect to \(y\).

Key Concepts

Area between curves

Finding the area between curves is a common task in calculus and it often involves integration. When you want to calculate the area between two curves, you're essentially finding the area of the region that's enclosed by them. Let's break this down further:

• Start by finding the points of intersection. These points give you the limits of integration as they define where the curves meet.
• Once you have these limits, determine which curve is on top and which is on the bottom over the interval you're considering. This is crucial because to find the area, you need to compute the integral of the top curve minus the bottom curve.
• This subtraction will give you the area between the two curves as opposed to the area under only one curve.

Using definite integrals, you can calculate the exact area bounded by these curves without relying on geometric approximations.
In the exercise above, to find the area between the given curves by integrating with respect to either variable, understanding how each curve behaves in relation to the other over the interval is key.

Definite integral

Definite integrals are a fundamental tool in calculus and are used to calculate areas under curves. In contrast to indefinite integrals, which represent a general form of antiderivatives, definite integrals give a specific numerical value that represents the "accumulated quantity," such as area under a curve from one point to another.

• The process begins by identifying an interval along the x-axis or y-axis over which to integrate. This interval is defined by two limits, the lower and upper limits.
• Once you have your interval, you integrate the function over this interval, which will yield a number.
• This number corresponds to the total area under the curve within the specified limits.

In the exercise provided, integrating with respect to both variables, x and y, shows how definite integrals can be applied in different ways to find the area of a region bounded by two curves.
Remember, whether you integrate with respect to x or y, the interpretation remains similar: calculating a cumulative value that represents a physical quantity, such as area, between the specified boundaries.

Quadratic equation

Quadratic equations play a pivotal role in finding intersections of curves, especially when these curves can be represented by parabolic shapes.
A typical quadratic equation takes the form:\[ax^2 + bx + c = 0\]

• To solve this equation, you can factor it, complete the square, or use the quadratic formula. Factoring is often the simplest (when possible), transforming the equation into a product of two binomials like \(y-p\)(y-q)=0\, where \(p\) and \(q\) are roots.
• In the provided solution, solving the quadratic equation \(y^2 + y - 6 = 0\) helped identify the points where both given curves intersect, yielding solutions \(y = 2\) and \(y = -3\).
• These solutions established the limits for the definite integrals used to find the area between the curves.

Quadratic equations are not only vital in analytics but are also heavily utilized in modeling various real-world phenomena, making them an essential topic in algebra and calculus.