Question

In Exercises \(15-34,\) find the area of the regions enclosed by the lines and curves. $$x+y^{2}=3 \quad$$ and $$\quad 4 x+y^{2}=0$$

Short answer

The area of the region enclosed by the curves is 16 units.

Step-by-step solution

Express the equations in terms of y.

Begin by solving each equation for x. This results in the following equations: \[x_1 = 3 - y^{2}\] and \[x_2 = -\frac{1}{4}y^{2}\] where \(x_1\) is for \(x+y^{2}=3\) and \(x_2\) is for \(4x+y^{2}=0\]

Find the intersection points of the two curves

Set the two equations equal to each other to find the points where these two curves intersect. \[3 - y^{2} = -\frac{1}{4}y^{2} \] Solving the above equation you get, \(y = -2\) and \(y = 2\)

Set up the integral

Now that you have the bounds and the functions, you can set up the integral. The area A of the region enclosed by the curves \(y=f(x)\) and \(y=g(x)\) from \(a\) to \(b\) is given by \[A = \int_{a}^{b} [f(x) - g(x)]dx\] Here our \(f(x) = 3 - y^{2}\), \(g(x) = -\frac{1}{4}y^{2}\), and \(a = -2\), \(b = 2\), so, \[A = \int_{-2}^{2} [(3-y^{2}) - (-\frac{1}{4}y^{2})]dy\]

Evaluate the integral

To find the area, now just compute the integral. \[A = \left. [3y-\frac{1}{3}y^{3}+\frac{1}{16}y^{3}]\right|_{-2}^{2}\] simplifying this gives you \[A = 16\]

Key concepts

Integration

Integration is a fundamental concept in calculus that is used to find the area under a curve, among other things. It starts with a function, like the one you have when you're dealing with the curves on a graph. To integrate a function means to find the integral, which represents a multitude of areas under the curve of the function. This process can be thought of as adding up an infinite number of infinitely small rectangles under the curve. In your exercise, integration was applied to find the area between two curves, which is a common application of this mathematical tool.

When calculating the area between curves, you construct an integral using the functions that describe these curves. The integral has bounds, which are the limits between which you want to calculate the area, and in this example, they are determined by the intersection points of the curves. Think of integration as a way to meticulously tally up the space between the lines, where every piece of the area counts, no matter how small.

Algebraic Manipulation

Algebraic manipulation refers to the various ways in which algebraic expressions and equations can be rearranged and simplified. This skill is crucial when setting up an integral for finding the area between curves. Initially, you start with the given equations and manipulate them to express one variable in terms of another. This was done in Step 1 of the solution, where both given equations were rearranged to solve for x.

After developing expressions for x in terms of y (or vice versa), these equations can be set equal to each other to find intersection points. This process involves moving terms from one side of an equation to the other, combining like terms, and factoring or expanding expressions. The ability to manipulate algebraic expressions lays the groundwork for setting up the correct integral for your area calculation.

Area Calculations

When it comes to area calculations between curves, the concept extends the basic area finding strategies to more complex shapes bounded by algebraic curves. As you've noticed from the problem, typically, each curve is described by an equation that relates the x and y coordinates of points on that curve. The area between these curves is a region on the graph; it's like finding the space inside a racetrack where each curve acts like a fence.

The general approach to finding this area is to take the integral of the 'top' function subtracted by the 'bottom' function between the bounds given by their intersections. However, remember, the terms 'top' and 'bottom' are in relation to how the curves are positioned on a standard graph, where the 'top' curve is the one with the higher y-value in the region of interest. The solution carries out these steps methodically, ensuring that the area enclosed is accurately calculated by taking advantage of the definite integral that spans from one boundary to the other.

Definite Integrals

Definite integrals are integrals with specific starting and ending points, called the lower and upper limits, respectively. These are the bread and butter of finding areas under curves and, in this context, between curves. A definite integral provides a precise numerical value, as opposed to indefinite integrals, which represent a family of functions.

In the problem you are working on, the definite integral was used to calculate the exact area between the curves from the leftmost intersection point to the rightmost one. The integral was defined with lower and upper bounds ((-2) and (2), correlating to the intersection points of the curves when y was set to (-2) and (2). Evaluating the definite integral, through techniques like the fundamental theorem of calculus, gives you a concrete number that represents the area. This is captured in the final answer of your solution, where evaluating the integral from one limit to the other resulted in the area of 16 square units.