Use the result of Exercise 17 to show that 0 ⩽∬R sinπx cosπy d A ⩽(1)/(32) where R=[0, (1)/(4)]

Use the result of Exercise 17 to show that 0 ⩽∬R sinπx cosπy...

Key Concepts

Double Integrals

Double integrals allow us to compute the accumulation of a function over a two-dimensional region by integrating over each variable in succession. This fundamental concept in multivariable calculus enables the evaluation of quantities such as area, volume, and mass by considering the contributions of infinitesimal elements across a region.

Iterated Integration (Fubini's Theorem)

Fubini’s Theorem justifies the process of evaluating double integrals as iterated integrals when the function is integrable over the rectangular region. By transforming a double integral into a sequential process of one-dimensional integrations, one can compute complicated integrals more easily using known antiderivatives and integration techniques.

Integration of Trigonometric Functions

Integrating trigonometric functions such as sine and cosine over specified intervals is a common application in calculus. Understanding the properties and periodic behavior of these functions is crucial, as it allows for the computation of definite integrals and simplifies the analysis of integrals that involve oscillatory behavior.

Establishing Integration Bounds (Integral Inequalities)

Establishing bounds for an integral involves estimating the integral’s value by comparing it to known quantities or employing inequalities that limit its possible size. This technique is useful for showing that an integral lies within a specific range, particularly when direct evaluation is complex or when previous results can provide tighter estimates.

Transcript

00:01 So this problem makes use of the results of a previous problem in order to solve it.

00:06 So i have boxed the results of the previous problem, which we shall be using in this problem.

00:12 And the function we are given, our f of x, y, i will be using blue for x and green for y for this problem.

00:23 And our f of xy is sine pi x, cosine pi y, cosign pi y.

00:35 So this problem is asking us to verify the maximum and minimum values of the double integral of this function over a certain region r, and they give us numbers, but we'll get to that later.

00:45 And before we go into any calculus, let's just take a closer look at this function by zooming in on its individual parts.

00:53 So let's look at the x part, which is sine pi x.

00:57 Okay? actually, we need the region right now.

01:05 So the region they give us in the problem, 0 to 1 quarter for the x part, and then 1 quarter to 1 half for the y part.

01:18 So for sine pi x over the interval of 0 to 1 over 4.

01:29 So if we take a look at the graph of sine pi x, so at x equals 1 half, this is sign over 2, which is 1.

01:39 So that means that x equals 1 over 4 is somewhere around here, and that value is here.

01:46 So our interval is from x equals 0 to x equals 1 over 4, which is between here and here.

01:54 So over this region, the highest value occurs at x equals 1 over 4, and the lowest value occurs at x equals 0.

02:02 So the minimum is sine, pi x equals 0 which is sine 0 equals 0 and our maximum occurs at x equals 1 over 4 so sign pi 1 over 4 which is sign pi over 4 which is a special one and that's equal to 1 over root 2 if we move on to the y portion where the function is cosine pi y our interval is from 1 quarter to one half.

02:41 And so we take a look at the graph of cosine.

02:47 So it's going to go like this and it's going to keep going to because it's sinusoidal.

02:52 This is going to be one, it's going to be zero at pi over two.

02:57 This is cosine.

02:59 Pi y, meaning that a quarter is somewhere around here.

03:05 So over this interval, the highest point, which is the max, occurs at x equals 1 over 4.

03:18 This is x, by the way.

03:20 It's a cosine pi 1 over 4, which is a special one, which is 1 over 2.

03:26 And the minimum over this interval occurs at x equals 1⁄2.

03:33 So, cosine pi over 2, which is equal to 0...