Question

Set up a double integral to find the volume of the solid bounded by the graphs of the equations. \(z=x y, z=0, y=x, x=1,\) first octant

Short answer

The double integral to find the volume of the solid bounded by the mentioned equations is \(\int_{0}^{1} \int_{0}^{x} xy \, dy \, dx\).

Step-by-step solution

Set the limits for y

The first step is to determine the limits for y. In the first octant, y, x, and z are all non-negative. This means that y ranges from 0 to x, since y is equal to x.

Set the limits for x

The next step is to determine the limits for x. Here, x ranges from 0 to 1, as per the given boundaries in the problem.

Setup the double integral

Taking Step 1 and 2 into account, the double integral can be set up as follows: \(\int_{0}^{1} \int_{0}^{x} xy \, dy \, dx\) which represents the volume of the solid in question.

Key concepts

Volume Using Double Integrals

Finding the volume of a solid region using double integrals is a fundamental application in calculus. Double integrals extend the concept of a single integral to two dimensions to calculate the volume under a surface over some region. The significant point is that the volume of a solid bounded by a surface can be found by integrating the height of the solid above each point in the region of the xy-plane.

The double integral \(\int \int_R f(x, y) \, dx \, dy\) calculates the volume of the solid that lies above the region R and below the surface given by the function f(x, y). To make sense of the integral, we divide the region R into smaller rectangles, approximate the volume above each rectangle, and sum these volumes together. As these rectangles get smaller and more numerous, the sum approaches the exact volume of the solid.

In the provided exercise, we used the double integral to find the volume of the solid bounded by the graphs \(z=xy\), \(z=0\), and the lines \(y=x\) and \(x=1\), in the first octant. By setting up the integral \(\int_{0}^{1} \int_{0}^{x} xy \, dy \, dx\), we are summing the 'slices' of volume under the surface \(z=xy\) over the triangular region in the xy-plane bounded by \(y=0\), \(y=x\), and \(x=1\).

Triple Integral Calculus

Triple integral calculus generalizes double integral calculus to three dimensions. It allows for the calculation of volumes of more complex regions in space, as well as applications in physics such as calculating mass, density, and moments of inertia. The triple integral \(\int \int \int_E f(x, y, z) \, dx \, dy \, dz\) sums the value of a function f(x, y, z) over a three-dimensional region E.

The process involves integrating in one variable while treating the other two variables as constants, and then repeating this process for the other variables. Essentially, we are accumulating the infinitesimal volumes 'dv' represented by \(f(x, y, z) \, dx \, dy \, dz\) throughout the space E.

While our current problem involves a double integral—suitable for calculating volumes under a surface—a triple integral would be used if we were interested in, for example, the mass of the solid with a given density function that varies by position.

Integration Limits Setup

Setting up integration limits is a critical step in performing multiple integrals. As seen in the exercise, we determine the limits of integration for each variable by examining the geometry of the region over which we're integrating. For integration over a planar region, we commonly set up limits for x and y.

To set the limits for the y-variable, we observe how y behaves relative to x within the region of interest: the lower limit is where y begins, and the upper limit is where y ends. In the provided exercise, y starts at 0 and increases up to the line y=x, so the limits for y are from 0 to x.

Then we establish the x-limits by looking at the 'leftmost' and 'rightmost' boundaries of the projected region on the x-axis. In our case, x varies from 0 to 1. Once the limits are set, the double integral can be written with the correct bounds, ensuring that we are integrating over the correct area to find the desired volume or measure.

When setting up integration limits for triple integrals, a similar approach is taken by considering the region's boundaries in all three dimensions. For every specific scenario, the structure of the region may require different strategies to determine the appropriate bounds for integration.