Question

Use double integrals to calculate the volume of the following regions. The solid beneath the cylinder \(z=y^{2}\) and above the region \(R=\\{(x, y): 0 \leq y \leq 1, y \leq x \leq 1\\}\)

Short answer

Answer: The volume of the solid is \(\frac{1}{9}\).

Step-by-step solution

Set up the double integral

First, we need to set up the double integral that represents the volume of the solid beneath the given cylinder and above the region R. Our double integral to compute the volume will have the following form: $$V = \iint_R z\,dA$$ In this case, the function z is given by \(y^2\). The domain R is the region defined by the given inequalities, which describes a triangle in the xy-plane. We will integrate over y first, and then over x. The limits of integration for y are from 0 to 1, and for x, they are from y to 1. $$V = \int_0^1 \int_y^1 y^2\, dx dy$$

Evaluate the inner integral

We will first evaluate the inner integral, which is a simple one-variable integral with respect to x. $$\int_y^1 y^2 dx = \left[\frac{1}{3}x^3y^2\right]_y^1 = \frac{1}{3}(1^3y^2-y^3y^2) = \frac{1}{3}(y^2-y^5)$$

Evaluate the outer integral

Now, we will evaluate the outer integral with respect to y, which also is a simple one-variable integral. $$V = \int_0^1 \frac{1}{3}(y^2-y^5) dy = \frac{1}{3}\left[\frac{1}{3}y^3 - \frac{1}{6} y^6\right]_0^1 = \frac{1}{3}\left[\frac{1}{3} - \frac{1}{6}\right] = \frac{1}{9}$$ So, the volume of the solid beneath the given cylinder and above the region R is \(\frac{1}{9}\).

Key concepts

Volume Calculation using Double Integrals

When we think of volume, we often imagine filling up a space with small cubes or another unit of measure. In calculus, especially with double integrals, the process involves summing up infinitely small volumes to cover a region. When dealing with solids that lie beneath a surface like a cylinder, double integrals allow us to compute the entire volume by integrating over a specified region.

In this problem, the volume is determined by setting up a double integral over the region of integration, denoted as R, beneath the surface described by the equation \( z = y^2 \). Here, the equation describes how high above the xy-plane the solid reaches at any point \( y \). The overall volume \( V \) can be expressed as:
\[ V = \iint_R z \, dA \] where \( dA \) represents an area element in the xy-plane.

Understanding the Cylinder in the Problem

In this context, a cylinder isn't just a typical 3D shape you might draw in school. It is better thought of as a surface or collection of lines. For this exercise, the cylinder is defined by \( z = y^2 \).

This equation implies that for every point \( (x, y) \) in the region, the height of the cylinder above the xy-plane is \( y^2 \). As \( y \) changes, the height of the cylinder changes, effectively creating a curved surface. This makes it quite unique compared to the traditional cylinders that are uniform across their height.

Therefore, the focus is on slicing the cylinder horizontally due to the variation of \( y \), and the maximum height of this cylinder at any point in our region of integration matters for our volume calculation.

Defining the Region of Integration

The region of integration specifies the part of the xy-plane over which the solid stretches. This problem defines the region R as a set of points where

• \(0 \leq y \leq 1\)
• \(y \leq x \leq 1\)

Graphically, this describes a triangular region. It begins at the origin (0, 0), stretches vertically to (0, 1), and moves horizontally to (1, 1). Within this triangle, each set of points \((x, y)\) serves as a base point from where the cylinder \(z = y^2\) rises.

Understanding this region is crucial because it directly impacts the evaluation of the double integral since it determines the limits of integration.

Setting and Evaluating the Limits of Integration

The limits of integration are boundaries that define the region to evaluate within the double integral. They are crucial for accurate volume computation.

For our problem, integrating over \( y \) first from 0 to 1, allows us to follow the line \(x = y\) to \(x = 1\). The limits of integration are set as follows:

• For \( y \) from 0 to 1, representing the vertical span.
• For \( x \) from \( y \) to 1, representing the horizontal distance covered at each \( y \).

These limits carve out the triangular region in the xy-plane over which the integration is done.

Each of these limits performs a pivotal role in determining the result of the integration, ensuring that calculations cover precisely the space of interest without including any extra areas.