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EnteredAnswer PreviewResult42.53322x(8 sin(8) + cos(8) _ 1)incorrectThe answer above is NOT correct:point) Evaluate the double integralcos Vx? +YdA, where D is the ...

Question

EnteredAnswer PreviewResult42.53322x(8 sin(8) + cos(8) _ 1)incorrectThe answer above is NOT correct:point) Evaluate the double integralcos Vx? +YdA, where D is the disc with center the origin and radius 2, by changing topolar coordinates Answer: 2pij8sin(8)+cos(8)-1]

Entered Answer Preview Result 42.5332 2x(8 sin(8) + cos(8) _ 1) incorrect The answer above is NOT correct: point) Evaluate the double integral cos Vx? +YdA, where D is the disc with center the origin and radius 2, by changing to polar coordinates Answer: 2pij8sin(8)+cos(8)-1]



Answers

$7-14$ Evaluate the given integral by changing to polar coordinates.
$\iint_{D} \cos \sqrt{x^{2}+y^{2}} d A,$ where $D$ is the disk with center the
origin and radius 2

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.

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.

So here on this problem we have to evaluate the integral by changing into polar connects. And the government integral is double integration over the region. D. X. Squared Y. D. A. This D. Is the region of top half of the disk with center at the origin And the radius is five. So this is this shaded region is D. We have to evaluate this integral dysfunction on this region now. And the rules changed into the polar cornets. So first it will be changing this into polar cornets. So for polar cornets we have X. Equals two arc assign theater. And why is equals to our science data and D. A. S. R. D. R. D. Tito. So this in trouble would become the double integration X. Square means are square science material times. Why? So why is our scientist? Um So this this R. And this are will become our cube so we can write our cube. Scientist. Oh I hope you got it. Thanks to you. So D. A. We have all dear. Did you know? So this is are and these are and these are will become out of the power for serving in right R. To the power for hair. And this is D. R detail. Now next we have to put the limits for D. R. And D. T. To And for that we have to consider this region. D have received that are is wearing between 0-1. So the limit would be our is less than sorry, greater than or equal to zero and less than or equals 25. And theatre is wearing between 0-5. So we can write that tita is less than or equals greater than again A greater than or equal to zero or less than or equals two pi. So you can put the limit share 0-5 and data we have zero to pi. No we can easily evaluate this double in trouble. So It will be equals two. Now the integration of our to the power five. Out of the power four with respect to D. R. Would be out of the power five divide by five. And the school science coated in science culture is concerned. Soak a science square Tito times scientist. Oh this constant with because we're integrating with respect would be our first. So here it would become out of the power of five divide by five. And the limit is from 0 to 5. And this integration as it is 0 to Pi and that is dated air. I'm going to put the limits upper limit minus the lower limit. So let me put the limits 0 to pay. This is cosine squared theta science potato authorities. Scientist only. Now here it is. Out of the power of five so it would become five to the power five divide by five minus the lower limit which is zero. So no need to write so dated as it is. This will become five to the power four because this five will cancel out and well we can take five to the power for common so it would become one here. All we can reply dated now. So this is the expression that we got. We can further we will evaluate this with respect to this theater, not evaluate this interval. We're going to use substitution method and we're going to substitute geoscience data as T. And after the visiting the other side will get minus scientist. Oh digital is equal to D. T. So this integral would become this is called Zion square kilometers T square scientists are digital means minus duty so they can write minus DT here and five to the power for this is as it is. So we can write five to the power four and this minus negative sign as it is. Now. We have to carefully put the limits here. We have changed the variable so we have to change the limit also. So when you know because I am theater we have taken into BST. Now Theater was wearing between zero. Uh Pine. So when tita is zero So T would be co sign zero and co sign zero would mean one. So we can write one hair. now when uh chosen this 20 to become spy. So co signed by would be called to take and cosine pi. We know it is a negative one. So the liberty would be negative one. Now we have uh we're gonna integrate this uh easily -5 to the Power four as it is. And this would become T. Q. Divided by three. And the lower limit is one. Upper limit is negative one. Now you can put the limits, This is -5 to the bottom four, divide by three. And a common. Our Parliament we have ah minus one and minus one cube. Mine is a low limit which is one here. So one Cube Okay for to simplify this and we'll get our answer to be nearly 1250 divide battery. Therefore, we can conclude that the double in struggle over the region the X square Y. D. A. Equals 21,250 divided by three. So this require answer for the given problem. I hope you have understood problem. Thank you.


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