Problem 87 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

CONSTRUCTION A storage bin is to be constructed in the shape of the solid bounded above by the surface $$ z=20-x^{2}-y^{2} $$ below by the \(x y\) plane, and on the sides by the plane \(y=0\) and the parabolic cylinder \(y=4-x^{2}\), where \(x, y\), and \(z\) are in meters. Find the volume of the bin.

Step-by-Step Solution

Answer

The volume of the bin is 83.2 cubic meters.

Step 1 - Understand the Problem

Identify the given surfaces and boundaries: The top surface is bounded by the equation \( z = 20 - x^{2} - y^{2}\), the bottom surface is the xy-plane \(z = 0\), the side by the plane \(y = 0\), and the other side by the parabolic cylinder \(y = 4 - x^{2}\).

Step 2 - Set Up the Volume Integral

Use a double integral to compute the volume of the bin. The volume integral is given by: \( V = \int \int_{R} (20 - x^{2} - y^{2}) \ dA \). The region \(R\) is defined by \(0 \leq y \leq 4 - x^{2}\) and \(-2 \leq x \leq 2\).

Step 3 - Evaluate the Inner Integral

First, integrate with respect to \(y\): \( \int_{0}^{4-x^{2}} (20 - x^{2} - y^{2}) \ dy \). This simplifies to \( \left[ 20y - x^{2}y - \frac{y^{3}}{3} \right]_{0}^{4-x^{2}} \).

Step 4 - Substitute and Simplify

Substitute the limits 0 and \(4-x^{2}\) into the expression: \(20(4-x^{2}) - x^{2}(4-x^{2}) - \frac{(4-x^{2})^{3}}{3}\). This expands and simplifies to: \(80 - 24x^{2} + \frac{8x^{4}}{3}\).

Step 5 - Evaluate the Outer Integral

Integrate this simplified expression with respect to \(x\) over \(-2\leq x \leq 2\): \( \int_{-2}^{2} (80 - 24x^{2} + \frac{8x^{4}}{3}) \ dx \).

Step 6 - Integral Calculations

Compute the integral piece by piece: \( \int_{-2}^{2} 80 \ dx = 160 \) \( \int_{-2}^{2} 24x^{2} \ dx = 128 \) \( \int_{-2}^{2} \frac{8x^{4}}{3} \ dx = \frac{256}{5} \right \). Combine these results: \( 160 - 128 + \frac{256}{5} \).

Step 7 - Final Calculation

Simplify the combined result to find: \[ 160 - 128 + 51.2 = 83.2 \].

Key Concepts

Double Integral

A double integral is a way to integrate over a two-dimensional area. In the context of this problem, it allows us to compute the volume of a three-dimensional solid. To set up a double integral, we need to identify the region of integration and the function to be integrated. For our storage bin, the volume is given by the double integral of the function representing the height of the bin, which is the surface equation. So, we integrate the expression \(20 - x^{2} - y^{2}\) over the region defined by the parabolic cylinder and other boundaries.

It's important to set up the correct limits of integration. This will involve looking at the projections on the xy-plane and correctly identifying how x and y range within this region. In our problem, the region R is represented by the inequality \(0 \leq y \leq 4 - x^{2}\), and \(-2 \leq x \leq 2 \).

Parabolic Cylinder

A parabolic cylinder is a three-dimensional surface obtained by extending a parabola along a straight line perpendicular to the plane of the parabola. In this problem, the parabolic cylinder is defined by the equation \(y = 4 - x^{2}\).

This means that for any fixed value of x, y is determined by the equation of the parabola, and it fills out the values along the y-direction from \(-\text{infinity}\) to \(\text{infinity}\). However, the actual bound in our problem does restrict y between 0 and the line given by \(4 - x^{2}\). This cylinder forms one of the boundaries for our storage bin. Thus, it directly affects the limits of our double integral and helps in visualizing the volume we need to calculate.

Boundaries in Calculus

In calculus, boundaries are crucial to defining the limits of integration. For double integrals, boundaries help delineate the region over which we integrate our function. They can be horizontal lines, vertical lines, or more complex curves like parabolas and circles. In our case:

• The upper boundary is the surface \(z = 20 - x^{2} - y^{2}\).
• The lower boundary is the xy-plane, where z = 0.
• On one side, we have the plane y = 0, and on the other, the limit set by the parabolic cylinder, \(y = 4 - x^{2}\).

We need to project these boundaries onto the xy-plane to form the integral limits. This involves finding the intersection points and correctly interpreting the left and right boundaries for the region of integration.

Integral Evaluation

Evaluating integrals, especially double integrals, can seem daunting, but it becomes simpler when broken into smaller steps:

1. Inner Integral: First, we integrate with respect to y. This determines the contribution to the volume from a vertical slice of the region.\[ \int_{0}^{4-x^{2}} (20 - x^{2} - y^{2}) \ dy \]

2. Substitute Limits: Substituting the limits into the result of the inner integral simplifies the expression.\[ 20(4-x^{2}) - x^{2}(4-x^{2}) - \frac{(4 - x^{2})^{3}}{3} \]

3. Outer Integral: Next, we integrate with respect to x. This computes the total volume across all vertical slices.\[ \int_{-2}^{2} (80 - 24x^{2} + \frac{8x^{4}}{3}) \ dx \]

Breaking it down this way helps in managing smaller, simpler calculations and combining them in the end.