Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the solid whose volume is the indicated iterated integral. \(\int_{0}^{1} \int_{0}^{1}(2-x-y) d y d x\)

Step-by-Step Solution

Answer

A tetrahedron bounded by \(z=2-x-y\) over the unit square \([0,1]\times[0,1]\).

Step 1: Understand the Integral Limits

Examine the limits of the integral. For the integral \( \int_{0}^{1} \int_{0}^{1}(2-x-y) \, d y \, d x \), the outer integral runs from \(x = 0 \) to \(x = 1\) and the inner integral runs from \(y = 0\) to \(y = 1\). This suggests a region in the \(xy\)-plane: a unit square with vertices at \((0,0)\), \((1,0)\), \((1,1)\), and \((0,1)\).

Step 2: Analyze the Function Inside the Integral

The function \(2-x-y\) defines the height of the solid at any point \((x, y)\) within the given domain. This is a plane in three-dimensional space, descending as \(x\) or \(y\) increases.

Step 3: Visualize the Overall Shape

Consider the solid formed. At \((x, y) = (0, 0)\), \(2-x-y = 2\). At \((x, y) = (1, 0)\), \(2-x-y = 1\). Similarly, at \((x, y) = (0, 1)\), \(2-x-y = 1\), and finally, at \((x, y) = (1, 1)\), \(2-x-y = 0\). The function defines a plane intersecting the \(z\)-axis at \(z=2\) and other endpoints described, creating a tetrahedron.

Key Concepts

Volume of a Solid

To comprehend iterated integrals, understanding how they relate to volume is crucial. An iterated integral essentially calculates the volume of a solid defined by a specific region. It involves stacking up infinitely many sheets in a manner prescribed by the function. Here, the integral \( \int_{0}^{1} \int_{0}^{1}(2-x-y) \, d y \, d x \) represents the volume under the surface \(2-x-y\) over the unit square on the \(xy\)-plane.- Each sheet corresponds to a slice of constant \(y\) or \(x\) and is integrated accordingly.- By integrating over the area where \(x\) and \(y\) go from 0 to 1, we're finding the accumulation of these sheets.Think of slicing through a loaf of bread where each slice represents the height at each point \((x,y)\), and the integral sums these to find the total volume. This is helpful in solving real-world problems in physics and engineering, where determining such volumes is necessary.

Three-Dimensional Space

Three-dimensional space is the realm where shapes like spheres, cubes, and in our case, a tetrahedron, exist. It consists of three axes: \(x\), \(y\), and \(z\). The function \(2-x-y\) defines a plane in this space.- The plane intersects the \(z\)-axis, which represents height, allowing us to visualize how the solid forms.- As you move along the \(x\)-axis or the \(y\)-axis, the height defined by \(2-x-y\) changes linearly, decreasing as you move farther from the origin.Understanding how points on the \(xy\)-plane translate into elevations on the plane \(z = 2-x-y\) is vital. It show's how the solid's shape emerges in three dimensions, extending beyond simple 2D shapes.

XY-Plane

The \(xy\)-plane is a two-dimensional slice of the three-dimensional space where \(z = 0\). In this context, it is pivotal as the domain onto which the integral is projected.- The limits \(0\) to \(1\) for both \(x\) and \(y\) define a square in this plane, delineating the base of our solid.- This square has vertices at \((0,0)\), \((1,0)\), \((1,1)\), and \((0,1)\), framing the area over which the height function, \(2-x-y\), acts.Visualizing this plane as a flat surface gives insight into integration limits and the region they cover. The area on this plane is essentially the canvas upon which the function paints its structure, providing the shadow or footprint of the solid in three-dimensional space.

Limits of Integration

The limits of integration determine the boundaries over which an integral is computed. They tell you where the integration starts and ends for each variable.- For the integral \( \int_{0}^{1} \int_{0}^{1}(2-x-y) \, d y \, d x \), the outer limits \(x = 0\) to \(x = 1\) define the range along the \(x\)-axis.- Similarly, the inner limits \(y = 0\) to \(y = 1\) describe the range along the \(y\)-axis.These boundaries help slice the problem into manageable pieces, letting you calculate the total volume by adding up contributions from each slice. Reading the limits is akin to defining the edges of a box within which you want to find what lies inside, helping to confine and pinpoint the solution in both mathematical problems and physical interpretations.