Problem 63 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Find the volume of the solid under the surface \(z=f(x, y)\) and over the given region \(R\).\(f(x, y)=x+1 ; R\) is bounded by \(y=8-x^{2}\) and \(y=x^{2}\).

Step-by-Step Solution

Answer

The volume is 16 cubic units.

Step 1 - Understand the region R

The region R is the area in the xy-plane bounded by the curves: \(y=8-x^2\) and \(y=x^2\).

Step 2 - Identify the points of intersection

To find the points of intersection, set \(y=8-x^2\) equal to \(y=x^2\): \[ 8 - x^2 = x^2 \] This simplifies to: \[ 8 = 2x^2 \] \[ x^2 = 4 \] Thus, \(x = 2\) or \(x = -2\). The points of intersection are (-2, 4) and (2, 4).

Step 3 - Set up the double integral

The volume under the surface \(z = f(x, y) = x + 1\) over the region R can be represented by the double integral: \[ \text{Volume} = \int_{a}^{b} \int_{g(x)}^{h(x)} f(x, y) \,dy \,dx \] Here, \(a = -2\), \(b = 2\), \(g(x) = x^2\), and \(h(x) = 8 - x^2\).

Step 4 - Integrate with respect to y

First, integrate \(f(x, y) = x + 1\) with respect to \(y\): \[ \int_{x^2}^{8-x^2} (x + 1) \,dy \] This simplifies to: \[ (x + 1) \left[y\right]_{x^2}^{8-x^2} \] \[ = (x + 1) \left[(8 - x^2) - x^2\right] \] \[ = (x + 1) \left[8 - 2x^2\right] \] \[ = 8x + 8 - 2x^3 - 2x^2 \]

Step 5 - Integrate with respect to x

Now integrate the resulting expression with respect to \(x\) from \(-2\) to \(2\): \[ \int_{-2}^{2} (8x + 8 - 2x^3 - 2x^2) \,dx \] This can be split into separate integrals: \[ \int_{-2}^{2} 8x \,dx + \int_{-2}^{2} 8 \,dx - \int_{-2}^{2} 2x^3 \,dx - \int_{-2}^{2} 2x^2 \,dx \] Evaluate each one separately: First, \( \int_{-2}^{2} 8x \,dx = 0\) because the integrand is an odd function. Second, \( \int_{-2}^{2} 8 \,dx = 8 (x\bigg|_{-2}^{2} = 32)\). Third, \( \int_{-2}^{2} 2x^3 \,dx = 0\) because the integrand is an odd function. Fourth, \( \int_{-2}^{2} 2x^2 \,dx = 2 \left[\frac{x^3}{3}\right]_{-2}^{2} = 16\). Adding these up, the result is: \[ 0 + 32 - 0 - 16 = 16 \]

Key Concepts

Double Integral

A double integral is a way to calculate the volume under a surface in a given region. It lets us add up tiny pieces of volume over a region in the xy-plane. We use double integrals, especially when the surface or the region below it is irregular. This exercise shows how to find the volume under the surface defined by the function \(z = f(x, y) = x + 1\) over a specified region \R\.
This is done using the formula: \[ \text{Volume} = \int_{a}^{b} \int_{g(x)}^{h(x)} f(x, y) \,dy \,dx\]
Here's a breakdown of the key parts:

• \(a\) and \(b\) are the limits for the outer integral, representing the range for \(x\).
• \(g(x)\) and \(h(x)\) are the limits for the inner integral, representing the range for \(y\) as functions of \(x\).
• \(f(x, y)\) represents the surface or function in question.

By evaluating this double integral, we determine the total volume under the given surface over the specified region.

Regions in the xy-plane

To set up a double integral correctly, we need to understand the region \R\ in the xy-plane. This region is bounded by two curves: \(y = 8 - x^2\) and \(y = x^2\).
The first step is to identify where these curves intersect. Setting them equal gives: \[ 8 - x^2 = x^2\]
Simplifying, we find \( x^2 = 4 \) or \( x = \pm 2 \). Therefore, the points of intersection are \(-2, 4\) and \(2, 4\).
These intersections tell us the horizontal range of \(x\) over which we integrate. Vertically, for each \(x\) in the interval \([-2, 2]\), \(y\) ranges from \(x^2\) to \(8 - x^2\). Understanding these bounds helps us set up the double integral and visualize the area we are working with.

Integration

Integration is the process of finding the whole from the sum of its parts. In the context of double integrals, we perform integration twice: once with respect to \(y\) and once with respect to \(x\).
For this problem, we first integrate \(f(x, y) = x + 1\) with respect to \(y\): \[ \int_{x^2}^{8-x^2} (x + 1) \,dy \ = (x + 1) [8 - x^2 - x^2] \ = (x + 1) [8 - 2x^2] \ = 8x + 8 - 2x^3 - 2x^2 \]
Next, we integrate the resulting expression with respect to \(x\) from \(-2\) to \(2\): \[ \int_{-2}^{2} (8x + 8 - 2x^3 - 2x^2) \,dx\]
We can break this into simpler parts and evaluate each:

• \ \int_{-2}^{2} 8x \,dx = 0 \ (since \(8x\) is an odd function)
• \ \int_{-2}^{2} 8 \,dx = 32
• \ \int_{-2}^{2} 2x^3 \,dx = 0 \ (since \(2x^3\) is an odd function)
• \ \int_{-2}^{2} 2x^2 \,dx = 16

Adding these results together gives us the final volume: \[0 + 32 - 0 - 16 = 16 \] This is the total volume under the surface \(z = x + 1\) over the region bounded by \(y = 8 - x^2\) and \(y = x^2\).