Question

Find the area of the region bounded by \(y^{2}+4 x=0\) and \(\left(\mathrm{y}^{2}+4\right) \mathrm{x}+8=0\).

Short answer

Answer: The area of the region is \(8\sqrt{2}\).

Step-by-step solution

Rewrite The Equations Explicitly

We first rewrite the given equations into explicit form: 1. \(y^2 + 4x = 0 \implies x = -\frac{y^2}{4}\) 2. \((y^2 + 4)x + 8 = 0 \implies x = -\frac{y^2}{4} - 2\) Now we have the explicit forms as: \(x_1 = -\frac{y^2}{4}\) and \(x_2 = -\frac{y^2}{4} - 2\).

Find Intersection Points

Next, we find the points where \(x_1 = x_2\). That is, where: \(-\frac{y^2}{4} = -\frac{y^2}{4} - 2\) Solving for \(y^2\), we get: \(y^2 = 8\) Thus, \(y = \pm 2\sqrt{2}\). To find the corresponding \(x\) values, we can use either of the expressions for \(x\). We'll use the first one: \(x = -\frac{y^2}{4} = -\frac{(2\sqrt{2})^2}{4} = -2\) So the intersection points are \((-2, -2\sqrt{2})\) and \((-2, 2\sqrt{2})\).

Set up the Integral for Area

We now set up the integral for the area between the two curves. Since we have the equations as a function of \(y\), we will integrate with respect to \(y\): $$\int_{-2\sqrt{2}}^{2\sqrt{2}} (x_2 - x_1) dy = \int_{-2\sqrt{2}}^{2\sqrt{2}} \left(-\frac{y^2}{4} - 2 - \left(-\frac{y^2}{4}\right)\right) dy$$ Simplifying, we get: $$\int_{-2\sqrt{2}}^{2\sqrt{2}} (-2) dy$$

Evaluate the Integral

Now we can evaluate the integral to find the area between the curves: $$-2 \left[\int_{-2\sqrt{2}}^{2\sqrt{2}} dy\right] = -2 [y]_{-2\sqrt{2}}^{2\sqrt{2}} = -2 (2\sqrt{2} -(- 2\sqrt{2})) = -8\sqrt{2}$$ Since the area cannot be negative, we take the absolute value to get the area: $$A = |{-8\sqrt{2}}| = 8\sqrt{2}$$ Therefore, the area of the region bounded by the given curves is \(8\sqrt{2}\).

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the accumulation of quantities and the areas under and between curves. When faced with a problem that requires finding the area bounded by curves, we use integrals to sum up infinitesimally small pieces of the region to get a finite answer.

In essence, integral calculus allows us to quantify the space within a region by taking the integral of a function. In our exercise, integral calculus is applied to determine the area between two parabolic curves by integrating the difference of their respective functions with respect to the variable of integration, in this case, 'y'. It's like adding up the 'slices' of area between the curves from one boundary to the other.

Area Between Curves

The area between curves is calculated by integrating the difference between the top function and the bottom function over the interval between their intersection points.

To establish this area, it's important to understand which curve is upper (or to the right when integrating with respect to 'y') and which one is lower (or to the left). The area is then the integral of 'top minus bottom' or 'right minus left', depending on the context. In our exercise, we determined that the area is found by subtracting the function for one curve from the function for the second curve and integrating this difference over the interval between their intersection points.

Intersection Points

Finding the intersection points of curves is crucial because it determines the interval over which we will integrate. Intersection points are where the graphs of two or more functions meet.

In the given exercise, we found the intersection points by setting the two functions equal to each other and solving for 'y', which gave us the bounds of integration. In practical terms, these points represent the 'start' and 'end' of the region we're interested in and serve as the limits of the definite integral for calculating the area.

Definite Integral

A definite integral has both a lower and an upper limit, which provide the bounds of the integration. It is used to calculate the total 'accumulation' of a quantity, often represented as the area under a curve between two points.

The integral symbol with limits, \[ \int_{a}^{b} f(x) dx \], means summing all infinitesimal areas under the curve 'f(x)' from 'x = a' to 'x = b'. In our example, the definite integral was used to determine the area of the region bounded by two specified values on the 'y'-axis, those being the intersection points previously determined. By evaluating this expression, we calculate the exact area enclosed by our curves within the specified limits.