Problem 42 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves, using (a) vertical elements and (b) horizontal elements. $$y=4 x, y=x^{3}$$

Step-by-Step Solution

Answer

The area between the curves is 16.

Step 1: Find points of intersection

First, we need to find the points where the curves intersect. Solve the equation \(4x = x^3\) by setting \(y\) of both equations equal to each other. Rearrange to get \(x^3 - 4x = 0\). Factor this equation to get \(x(x^2 - 4) = 0\). This gives us \(x=0\), \(x=2\), and \(x=-2\) as points of intersection.

Step 2: Set up integrals for vertical elements

Using vertical elements, the area between the curves from \(x = -2\) to \(x = 2\) is given by the integral \(\int_{-2}^{2} (4x - x^3) \, dx\). This expression considers that \(y = 4x\) is above \(y = x^3\) within these limits.

Step 3: Calculate integral for vertical elements

Compute the integral: \[\int_{-2}^{2} (4x - x^3) \, dx = \left[2x^2 - \frac{x^4}{4}\right]_{-2}^{2}\]. Evaluate this from \(-2\) to \(2\): \[\left(2(2)^2 - \frac{(2)^4}{4}\right) - \left(2(-2)^2 - \frac{(-2)^4}{4}\right)\]. Simplifying gives \(16\).

Step 4: Set up integrals for horizontal elements

Using horizontal elements, the area is split into two parts: from \(y = -8\) to \(y = 0\) where \(x = y^{1/3}\) is on the left of \(x = \frac{y}{4}\), and from \(y = 0\) to \(y = 8\) where \(x = \frac{y}{4}\) is on the left of \(x = y^{1/3}\). Thus, compute: \[\int_{-8}^{0} (y^{1/3} - \frac{y}{4}) \, dy + \int_{0}^{8} (\frac{y}{4} - y^{1/3}) \, dy\].

Step 5: Calculate integral for horizontal elements

Compute the integrals separately. For \(\int_{-8}^{0} (y^{1/3} - \frac{y}{4}) \, dy\), evaluate as \[\left[\frac{3}{4}y^{4/3} - \frac{y^2}{8}\right]_{-8}^{0}\]. For \(\int_{0}^{8} (\frac{y}{4} - y^{1/3}) \, dy\), evaluate as \[\left[\frac{y^2}{8} - \frac{3}{4}y^{4/3}\right]_{0}^{8}\]. Both calculations simplify to an area of \(16\).

Step 6: Conclusion: Total area

Thus, the area between the curves is \(16\) using both vertical and horizontal integrations. This verifies consistency and correctness in both methods.

Key Concepts

Understanding Points of Intersection

When tackling the problem of finding the area between curves, identifying the points of intersection is a critical first step. These points determine the boundaries of integration. In this exercise, you'll start by equating the two curve equations, which are given by: \( y = 4x \) and \( y = x^3 \).

By setting these equations equal, you establish a common \( y \) value and solve the equation \( 4x = x^3 \). Rearrange this to \( x^3 - 4x = 0 \), and factor it as \( x(x^2 - 4) = 0 \). Factorization helps us identify the solutions: \( x = 0 \), \( x = 2 \), and \( x = -2 \). These intersections, \((-2, -8)\), \( (0, 0)\), and \((2, 8)\), demarcate the values where the two curves meet on the coordinate plane.

Integrals in Finding Area Between Curves

Integrals are the mathematical tools we use to find the area between curves. After finding the points of intersection, the next step involves setting up these integrals correctly.

For vertical elements, you look at the area between two functions along the \( x \)-axis from \( x = -2 \) to \( x = 2 \), integrating the difference of the functions. This is expressed as \( \int_{-2}^{2} (4x - x^3) \, dx \). Similarly, using horizontal elements, we determine the area along the \( y \)-axis between \( y = -8 \) to \( y = 8 \). This involves two separate integrals: \( \int_{-8}^{0} (y^{1/3} - \frac{y}{4}) \, dy \) and \( \int_{0}^{8} (\frac{y}{4} - y^{1/3}) \, dy \). Evaluating these integrals confirms the consistency in calculating the area as \( 16 \). The usage of integrals allows finding the precise area even when curves bend and twist.

Vertical Elements and Integration

Using vertical elements means integrating with respect to \( x \). This approach works well when the functions can be expressed as \( y = f(x) \) over the specified range.

In this exercise, first consider the interval from \( x = -2 \) to \( x = 2 \), where \( y = 4x \) is the upper function and \( y = x^3 \) is the lower. The area calculation involves: \( \int_{-2}^{2} (4x - x^3) \, dx \). Compute this integral by finding the antiderivative, which is \( 2x^2 - \frac{x^4}{4} \). Evaluating this from \( x = -2 \) to \( x = 2 \) gives \( 16 \). The key takeaway: subtraction of functions within the integral represents the space between them.

Utilizing Horizontal Elements

Horizontal elements involve integrating with respect to \( y \). This is useful when functions are expressed as \( x = g(y) \).

In this exercise, calculate the area by splitting it into two regions based on the intersections at \( y = 0 \). From \( y = -8 \) to \( y = 0 \), \( x = y^{1/3} \) is to the left of \( x = \frac{y}{4} \). Thus, integrate \( \int_{-8}^{0} (y^{1/3} - \frac{y}{4}) \, dy \). Then, from \( y = 0 \) to \( y = 8 \), \( x = \frac{y}{4} \) is to the left of \( x = y^{1/3} \), so calculate \( \int_{0}^{8} (\frac{y}{4} - y^{1/3}) \, dy \). Both parts simplify to define the same area of \( 16 \). Using horizontal elements helps when functions swap their order of being to the left/right of each other.