Question

In Exercises \(15-34,\) find the area of the regions enclosed by the lines and curves. $$y^{2}-4 x=4 \quad$$ and $$\quad 4 x-y=16$$

Short answer

The total area enclosed by the two lines can be calculated by finding the absolute value of difference of the integrals of the two curves over limits defined by intersection points.

Step-by-step solution

Reformatting and Simplifying the Expressions

Starting with the first equation \(y^{2} - 4x = 4\), rearrange it to the form \(y^{2} = 4x + 4\), which can be further written as \(y = \sqrt{4x + 4}\). Doing the same for the equation \(4x - y = 16\), the reformatted equation would be \(y = 4x - 16\).

Locate Intersection Points

To find the intersection points of the two curves, set the two equations equal to each other, which results in \(\sqrt{4x + 4} = 4x - 16\). To solve for \(x\), square both sides and solve the resulting equation. The roots of the quadratic equation are the \(x\)-values of the intersection points.

Calculate the Enclosed Area

Once the limits of integration (or intersection points) are found, the area can be evaluated by computing the integral of the absolute difference of the two equations from the lower limit to the upper limit.

Key concepts

Integration

In calculus, integration is a fundamental operation that allows us to calculate areas under curves, among other things. When tasked with finding the area of an enclosed region between two curves, integration becomes an essential tool. You often integrate the difference between the two functions over the interval where the functions intersect.

The process begins by visualizing the region whose area we wish to calculate. Then, determine which of the two curves forms the upper boundary and which forms the lower boundary of the region over the interval in question. In the given problem, we integrate the absolute difference between the two equations to account for which function is on top and which is on the bottom and to ensure we get a positive value for the area. It's essential indeed to use the absolute value, as this will prevent subtracting negative areas, which can occur when the top function dips below the bottom function.

After setting up the integral, we evaluate it from the lower to the upper limits, which are the x-values of the intersection points. The result of this definite integral will give us the total area of the enclosed region. Integration helps us not just to find areas under single curves, but the complex areas that lie between curves, showcasing its powerful utility in calculus.

Quadratic Equations

When two curves intersect, the coordinates of their intersection points often satisfy both equations simultaneously. This intersection often leads to a quadratic equation when one of the curves is quadratic in nature, as in our example with the equation \(y^2 - 4x = 4\).

To solve a quadratic equation, you may rearrange it into a standard form, such as \(ax^2 + bx + c = 0\), and then use a variety of methods to find its solutions or roots. These methods include factoring, completing the square, or using the quadratic formula. The roots of the equation correspond to the x-values where the two curves intersect. In our problem, after squaring both sides of the equation to eliminate the square root, we obtain a quadratic equation that gives us the x-values of the intersection points by solving it.

Solving quadratic equations is a foundational skill in algebra, with vast applications in calculus, such as finding the intersection points which are required to set the bounds for integration, when calculating the area between curves.

Intersection Points

Intersection points are the coordinates where two or more graphs meet or cross each other. These points are significant because they often provide the boundaries within which areas, distances, and other quantities are to be calculated, especially in integration problems.

To find intersection points between two curves, we set their equations equal to each other and solve for the variables. In our exercise, we are dealing with the functions \(y = \textcolor{white}{\sqrt{4x + 4}}\) and \(y = 4x - 16\). By setting these equal to each other, we can find the points where the two curves intersect. It's noteworthy that if the resulting equation is a quadratic, as it is in our case, there may be two intersection points to consider, assuming they both lie within the region of interest—in other words, the visible portion of the graphs where the area is being sought.

These intersection points are critical in determining the limits of integration for calculating the area of the enclosed region. Without correctly identifying the intersection points, we might integrate over incorrect bounds, leading to inaccurate area calculations.