In Exercises 15-22, evaluate the double integral over the given region R ∬R y sin(x+y) d A, R

step 1 if we want to evaluate this integral over our region here. Notice, since our balance are just a

In Exercises 15-22, evaluate the double integral over the...

Key Concepts

Double Integrals

Double integrals allow us to compute the accumulation of a quantity (such as area, volume, mass, etc.) over a two-dimensional region by integrating a function of two variables. This technique extends the concept of single-variable integration to higher dimensions and is fundamental in multivariable calculus.

Iterated Integrals and Fubini's Theorem

Fubini's Theorem provides the foundation for evaluating double integrals as iterated integrals, which means breaking down a double integral into two successive single integrals. This theorem is especially useful when the region of integration is rectangular, allowing for simpler and more systematic computation.

Integration over Rectangular Regions

When the region of integration is rectangular, the limits for each variable are constant and independent of the other variable. This simplifies the evaluation process because the order of integration can be chosen arbitrarily, and the integral can be computed as the product of the integrals over the separate intervals.

Integration Techniques (Substitution Method)

When the integrand is a composite function, such as a sine function of a sum of variables, the substitution method can be used to simplify the integration process. Identifying the inner function and its derivative helps in changing variables, making the integral more manageable.

Transcript

00:01 If we want to evaluate this integral over our region here.

00:07 Notice, since our balance are just a rectangular region, where x and y do not depend on each other, we can go ahead and integrate this in any order we want.

00:16 I'm going to go ahead and integrate y first, since i believe it will make it a little bit easier to do, but both ways you should be able to do this with no problem.

00:27 So we're going to put x on the outside, so it will be negative pi to zero, and then 0 to pi, y, sine of x plus y, d, y, d, x.

00:40 Now, on the inside here, we assume that x is a constant with respect to y.

00:49 Meaning when we go ahead and integrate this, we can first use integration by parts.

00:56 And i'm just going to use the diy method to make this a little bit faster.

01:00 So i want to take the derivative of y because that will eventually become zero and integrate sine of x plus y.

01:09 So y, sine of x plus y.

01:14 So this is going to be one and then zero, and then we're going to integrate the same amount of time.

01:19 So integrating this one time is going to be negative cosine of x plus y, and integrating this again will be negative sine of x plus y.

01:30 And now over on the side we do plus minus plus and our first term so it's going to be this diagonal multiplied together so actually let me write this out of here so negative pi zero and we're going to have negative y cosine of x plus y and then our next term is going to be negative one times negative of x plus y, which should just be positive, so plus sine of x plus y.

02:13 And then lastly, we would multiply this last row right here.

02:23 And we would still be integrating that.

02:25 So it would be minus in this case, because we have the negative there...

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