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Evaluate the double integral [I.V+++1 dA where D is quarter disk o center 0(0.0) aud of radius 2 What is the value of this integral O the other three quarters of th...

Question

Evaluate the double integral [I.V+++1 dA where D is quarter disk o center 0(0.0) aud of radius 2 What is the value of this integral O the other three quarters of the disc D. Justily YOlr answer.

Evaluate the double integral [I.V+++1 dA where D is quarter disk o center 0(0.0) aud of radius 2 What is the value of this integral O the other three quarters of the disc D. Justily YOlr answer.



Answers

Evaluate the integral.
$\int_{R} \sin \left(x^{2}+y^{2}\right) d A,$ where $R$ is the disk of radius 2 centered at the origin.

That this disk centered at the origin with radius to someone. We have a disk. We know our data's air running from zero two team pie and our radius. That's just your too All right. So let's change this into little into an integral and poor record in its sonar theater's air running from Syria to pie, our radio are running from zero to two, and we know this is going to be co sign of our PR later. Okay, again, I'm just using the fact that our squared is equal to X squared plus y squared so are just equal to the square root of that. Okay, so to solve this into girl Houdini's integration by plants Okay, so you was going to be equal toe are Devi is going to be co sign. Are D R D'Oh d'Oh and Veer's sign are just kind of are right. So this is integral is your day to die? Uh, our sign R from zero to minus the integral from zero to to sign her D'Or. All right, so those zero to two. And when we evaluate, we get to sign two minus zero plus co sign of our new zero to two. Nothing deserted too high. Those two sign two plus co sign to nine is co sign a zero. Because I have zero is just one. Okay. So that our dictators you see that this is completely independent of data. So this is a really easy integration for us to do. We just get to pine. Let's go sign with all this stuff there.

Do this with a calculator. So the first thing that we need to know is that we've got a disc radius one. So our theater is going to run from 0 to 2 pi r r is gonna run from 0 to 1 and are integral. Yeah, looks like this e r squared, squared R d r dictator. Okay, so the integral that we're evaluating just to write it nicely is for are to the fourth d r. Do you think, uh, and plugging this into a calculator? We get that as our answer.

Okay, We want to integrate this function. Cosine x squared plus y squared over this region which is a circle or a disk of radius to center +00 So let's go ahead and changed you Polar X squared plus y squared is R squared and then d a is r D r d theta are is going from here r equals zero Up to here are equals two and it's a whole circle. So from 30 equals zero all the way around 2 30 equals two pi. All right, we're gonna let you equal r squared then do you? Is to our do your And if r equals zero nu equals zero If r equals two you calls for so this integral changes to 0 to 2 pi 04 cause I knew do you d theta So the integral of the coastline of you is the here too, But by co sign, uh, sign you from 0 to 4 d theta. So is here to two pi Sign of four minus sign of zero. Do you theta? I can remember for his radiance, Zeros, Radiance The sign of zero radiant zero. So now we have sign of four 0 to 2 pi d theta. So that sign of four times data from 0 to 2 pi So that is two pi sign up four.

Were given the integral of a function over region, were asked to use polar coordinates to evaluate this integral. The integral is the double winds girl over the region de of co sign square root of exports y square d A where region de is the disc with center at the origin in a radius of two. So let's sketch a graph of this region. So we have the line data equals pi over two and they did equals zero. We see that this is going to be if I draw the region in red when you have a boundary of our equals two and it will contain all points in that boundary. So this red shaded region, this is the region de and in polar coordinates. This region is given by set of all our data such that are is going to lie between zero and two and data will lie between zero and to pipe square. Root function is defined as long as its argument is greater than or equal to zero. We have that X squared plus y square is always greater than or equal to zero so that the square of it is also defined and we have X squared is a continuous function. Why square his continuous function so X squared plus y squared is continuous. Therefore, it's where that is also continuous, and so the coastline of that is also continuous. So it follows that the function that call it F of X y equals co sign of square root of X squared plus y squared an F is continuous, including on the region de since our region de is a simple region and since our function is continuous on this region, it follows that the double integral over G of our function co sign of square root of X squared plus y squared D A is well defined and is given by formula in polar coordinates integral from 0 to 2 pi integral from zero to of co sign of square of and in polar coordinates. X squared plus y squared is R squared says is the square root of R squared which, because our is always positive, this is simply co sign or and then because of the change in variable to polar coordinates or DRG data using food beanies. The're, um because the bounds on the integral, our constant weaken right This is the product of two into girls and to go from 0 to 2 pi of deep data times the integral from 0 to 2 of our co sign of or d R first into girls Easy to evaluate. Second, Integral does not have a simple anti derivative, but we can use integration by parts to find the anti derivative. So we just look at this integral for a second. It's label you as so suppose that you was equal to our and D V is equal to co sign of our Dior. Therefore, you have that he is going to be equal to sign of our and d you is simply going to be d r and by integration by parts we have that the integral from 0 to 2 of our co sign of our d r is equal to you. Times the judges are sign are evaluated from you are too minus the integral from 0 to 2 of this was you DVDs and I were looking for V. D. U, which is simply sign our g r. This is equal to to sign to minus zero or to sign it too, minus taking anti derivatives negative co sign of ours. This is the same as plus co sign of our evaluated from It's Hard to, which is to sign up to plus co sign of to minus co sign of zero, which is one therefore returning to the original Integral we have that the integral from 0 to 2 pi a dif data taking a derivatives and solving is simply going to be two pi and then our second integral is equal to to sign too Waas co signing too minus one This is our answer.


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