Use the result of Exercise 41 to show that 0 ⩽∬R sinπx cosπy d A ⩽(1)/(32) where R=[0, (1)/(4)]

Use the result of Exercise 41 to show that 0 ⩽∬R sinπx cosπy...

Use the result of Exercise 41 to show that
$$0 \leqslant \iint_{R} \sin \pi x \cos \pi y d A \leqslant \frac{1}{32}$$
where $R=\left[0, \frac{1}{4}\right] \times\left[\frac{1}{4}, \frac{1}{2}\right]$

Key Concepts

Double Integrals

Double integrals allow the calculation of the accumulation of quantities (such as area, volume, or mass) over a two-dimensional region. They extend the concept of single-variable integration to functions of two variables, and when the region is rectangular, the integration can be performed as iterated integrals.

Fubini’s Theorem

Fubini’s Theorem asserts that if a function is integrable over a rectangular region, the double integral can be computed as an iterated integral by integrating with respect to one variable at a time. This separation simplifies evaluating the integral, especially when the function factors into a product of single-variable functions.

Integration of Trigonometric Functions

Integrating trigonometric functions such as sine and cosine involves using well-known antiderivative formulas and identities to simplify and compute the integrals. Their periodic behavior and established integral values over standard intervals are key in calculating or bounding such integrals.

Bounding of Integrals

Bounding an integral involves estimating its value by finding lower and upper limits based on the known bounds of the integrand. By determining the maximum and minimum values of a function over a given region, one can effectively constrain the range of the integral even when exact evaluation is complex.

Transcript

00:01 So what we'd like to do is use the results of exercise 41 to help prove our inequality.

00:06 So our inequality in this case is that the integral over r of sine pi x, cosine pi y, da, is in between the value 0 and 1 over 32.

00:15 Where r is our region, 0 to 1 4 in x and 1 4th to 1⁄2 in y.

00:21 So what we'd like to do is figure out where each side of the boundaries come from and prove them.

00:28 So i would say the zero portion is easier to prove, considering that we know that over the bounds 0 to 1 4th, sine of pi x is positive, so this is greater than 0.

00:43 And then cosine pi y is also greater than 0.

00:46 So since this is the integral of a positive value da over a region, we know that this integral must evaluate to a positive number.

00:57 So that's easy to prove.

01:01 Now what we need to prove is the fact that it's less than 1 over 32.

01:04 So we're giving the hint that we have to use exercise 41's result, and the results of exercise 41 is that an integral over, of a constant function over a rectangular region is just equal to this product.

01:19 So what we'd like to do is figure out how we can make a bound using a constant function.

01:25 So if we look at our trigonometric functions, we can realize that these over this region will have a maximum value.

01:39 So the maximum value that these can attain is based on what our bounds are.

01:47 So as sine pi x, x gets greater, sine pi x increases...

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