Problem 50 - Calculus Early Transcendentals

Question

Find the volume of the following solids. The solid beneath the paraboloid \(f(x, y)=12-x^{2}-2 y^{2}\) and above the region \(R=\\{(x, y): 1 \leq x \leq 2,0 \leq y \leq 1\\}\)

Step-by-Step Solution

Answer

Answer: \(\frac{23}{3}\) cubic units

Step 1: Write down the integral for the volume

To find the volume of the given solid, we will set up a double integral of the function f(x, y) over the region R as follows: $$V = \int\int_R f(x, y)dA$$

Step 2: Represent the function f(x, y)

The given function is: $$f(x, y) = 12 - x^2 - 2y^2$$

Step 3: Set up the limits of integration for the double integral

The given region R is stated as: $$R = \{(x, y): 1 \leq x \leq 2, 0 \leq y \leq 1\}$$ So, the limits of integration for x will be from 1 to 2 and for y will be from 0 to 1. Now, we can set up the double integral as follows: $$V = \int_{1}^{2} \int_{0}^{1} (12 - x^2 - 2y^2) dy dx$$

Step 4: Integrate with respect to y

First, we will integrate the inner integral with respect to y: $$\int_{0}^{1} (12 - x^2 - 2y^2) dy = 12y - x^2y - \frac{2}{3}y^3 \Big|_0^1$$ Now, substitute the limits for y: $$12 - x^2 - \frac{2}{3} = 10 - x^2 - \frac{2}{3}$$

Step 5: Integrate with respect to x

Now, we will integrate the outer integral with respect to x: $$V = \int_{1}^{2} (10 - x^2 - \frac{2}{3}) dx = 10x - \frac{1}{3}x^3 - \frac{1}{3}x \Big|_1^2$$ Now, substitute the limits for x: $$V = (20 - \frac{8}{3} - \frac{2}{3}) - (10 - \frac{1}{3} - \frac{1}{3}) = 10 - \frac{7}{3} = \frac{23}{3}$$

Step 6: State the final answer

The volume of the solid beneath the paraboloid \(f(x, y)=12-x^{2}-2 y^{2}\) and above the region \(R= \{(x, y): 1 \leq x \leq 2,0 \leq y \leq 1\}\) is \(\frac{23}{3}\) cubic units.

Key Concepts

Volume of Solids

To find the volume of a solid region in space, especially with varying surfaces on the top and bottom, we often use double integrals. This involves integrating a function that represents the solid's height over a defined region on the xy-plane. In this case, we have a function, the paraboloid surface denoted as \(f(x, y) = 12 - x^2 - 2y^2\), which represents the height of the solid above any point \((x, y)\) within the given limits of integration. The volume is essentially the sum of all infinitesimally small "columns" of solid, calculated by the integral:

• Inner integral: Integrate over one variable, holding the other constant to shape thin "slices" across one dimension.
• Outer integral: Sum these slices across the second dimension.

This process allows us to compute the entire 3D volume trapped under a surface, providing a powerful tool for analyzing complex geometries.

Paraboloid

A paraboloid is a three-dimensional surface that, in this problem, is defined by the function \(f(x, y) = 12 - x^2 - 2y^2\). It appears similar to a stretched bowl, opening downward here, where the highest point occurs at the center \((0, 0, 12)\). From this central peak, the paraboloid dips downward, decreasing in height as \(x\) and \(y\) move away from the center.

• Parabolic Sections: Cross-sections parallel to the xy-plane are ellipses, getting smaller as we move down the structure.
• Apex: The pinnacle of this surface offers the maximum value of the function, crucial for determining the constraint of volume.

Understanding the shape of the paraboloid helps visualize how the volume calculations cover all regions above or below a certain plane, making it essential in interpreting and carrying out the integration.

Limits of Integration

Limits of integration are crucial in setting up the bounds for a double integral, which helps to narrow down the region over which the solid's volume is being calculated. For this exercise, we consider the region \(R = \{(x, y): 1 \leq x \leq 2, 0 \leq y \leq 1 \}\). These constraints define a rectangular slice on the xy-plane.

• X-axis Limits: The bounds are determined by the interval \(1 \leq x \leq 2\), fixing the horizontal span.
• Y-axis Limits: The limits \(0 \leq y \leq 1\) form the vertical span which nest with the x-limits to outline the overall region over which the paraboloid extends.

Carefully applying these limits keeps the integral confined to the specified area, allowing us to compute the volume beneath the paraboloid only over that particular range, yielding the accurate geometry defined in the problem.