Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(x=4 y^{4}, x=8-4 y^{4}\)

Short answer

The area of the region is 6.4 square units.

Step-by-step solution

Understand the Functions

The two functions described by the equations are: \( x = 4y^4 \) and \( x = 8 - 4y^4 \). These represent two curves, with \( x \) described as a function of \( y \). The first curve starts at the origin and increases as \( y^4 \), and the second curve is a downward-facing curve mirrored around \( y \) and shifted upwards by \(8\).

Find Intersection Points

To sketch the region bounded by these equations, we first identify their intersection points. Set \( 4y^4 = 8 - 4y^4 \). Simplifying gives \( 8y^4 = 8 \), leading to \( y^4 = 1 \), so \( y = \pm 1 \). Thus, the intersection points of the curves occur at \( y = -1 \) and \( y = 1 \).

Sketch the Region

Draw the graphs of \( x = 4y^4 \) and \( x = 8 - 4y^4 \). The region of interest is bounded between the two curves, with intersections at \( y = -1 \) and \( y = 1 \). The bounded region resembles a 'lens' shape between these \( y \)-values and symmetric along the \( y \)-axis.

Choose a Typical Slice

Choose a horizontal slice (a typical rectangle) at an arbitrary \( y \) between \( -1 \) and \( 1 \). The width of the slice (along the \( x \)-axis) is the difference in \( x \)-values from the two equations: \((8 - 4y^4) - 4y^4 = 8 - 8y^4\).

Approximate the Slice Area

The area of a typical slice is its width times its height, \( dA = (8 - 8y^4) \, dy \), where \( dy \) is the infinitesimally small height of the rectangle.

Set Up the Integral

The total area is the sum of all infinitesimal slice areas from \( y = -1 \) to \( y = 1 \). This is evaluated by the integral: \[ A = \int_{-1}^{1} (8 - 8y^4) \, dy \]

Calculate the Integral

Calculate the integral: \[ A = \int_{-1}^{1} (8 - 8y^4) \, dy = \left[ 8y - \frac{8y^5}{5} \right]_{-1}^{1} \]. After substituting, we have: \[ A = \left( 8(1) - \frac{8(1)^5}{5} \right) - \left( 8(-1) - \frac{8(-1)^5}{5} \right) \] This simplifies to \( A = \left( 8 - \frac{8}{5} \right) - \left( -8 + \frac{8}{5} \right) \), which equals \( \frac{32}{5} \) or \( 6.4 \).

Verify with an Estimate

As a rough estimate, the region is approximately 2 units wide (intersection to intersection at x-axis) and 3 units tall (maximum \( x \) value \( 8 \)), giving an estimated area of around 6 units. Our calculated value of 6.4 is consistent with this rough estimate.

Key concepts

Area Between Curves

The concept of finding the area between curves is an essential component of calculus, especially when dealing with functions that intersect or form bounded regions. It involves calculating the space or region occupied between two curves within a specified domain of interest. This can be visualized as the net area from one curve to the other over an interval.
To find such an area:

• Identify the curves or functions on a given interval.
• Calculate the integral of the upper function minus the lower function within that interval.

In our exercise, the functions are set in terms of a variable that is typically not used, with both equations expressed as functions of \(y\). By finding the difference \((8 - 8y^4)\), we calculate just what lies between the curves, not beneath them. Calculus makes this procedure more streamlined and precise than estimating areas using geometrical shapes.
The true value of using integration is in its exactness, allowing us to perfectly measure the area between these curves without approximation.

Sketching Regions in Calculus

Sketching regions in calculus helps in visualizing the area we aim to compute. Understanding the sketch can significantly simplify the integration process, by offering insights into the behavior and intersection of the curves.
To draw the region:

• Plot both functions on a coordinate system. For this problem, you plot equations \(x = 4y^4\) and \(x = 8-4y^4\), both in a non-standard \(x-f(y)\) form.
• Mark the intersection points correctly, as they define the domain over which integration occurs. These are found where the curves' expressions equate, giving \(y = -1\) and \(y = 1\).
• Visually inspect the form or shape bounded by your curves. In the context of this problem, the shape is lens-like, visually confined by the vertices at points where curves intersect.

By laying out graphs accurately, you can confirm the positions and general size of the bounded area, setting the stage for integration to determine the area more precisely.

Definite Integrals

Definite integrals play an integral role in finding the area under a curve over a specific interval. They are used to compute the total area that falls between the curves in our problem.
The process involves:

• Setting up the integral expression with the integrand reflecting the difference between upper and lower curves.
• Evaluating this integral over the interval defined by the intersection points. In this exercise, the integral setup is \(\int_{-1}^{1} (8 - 8y^4) \, dy\).

The calculation computes the difference or net area between curves:\[A = \left[8y - \frac{8y^5}{5}\right]_{-1}^{1}\]This expression is evaluated at the upper boundary and compared against the lower boundary to get the requisite area of \(\frac{32}{5}\) or 6.4. Definite integrals offer a structured way to sum up infinitely small slices into a complete area, accurately and reliably which is key in any calculus-related problem.

Intersection Points of Curves

Intersection points are critical when determining the limits for integration or constructing bounds for regions between curves. They essentially pin down where one function meets another, marking the change in boundaries.
For the functions \(x = 4y^4\) and \(x = 8 - 4y^4\) in this problem, the intersection point evaluation comes from:

• Equating the two functions: \(4y^4 = 8 - 4y^4\).
• Simplifying results in \(8y^4 = 8\), which reduces to \(y^4 = 1\).
• Solving gives points where \(y = -1\) and \(y = 1\).

These points are both visually and analytically necessary. Visually, they help in drawing the correct boundary on a graph. Analytically, in integration, they cap the region over which you integrate and hence decide the computed area. Understanding where and how functions intersect provides clarity and ensures accurate area calculations between curves.