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Evaluate the iterated integral by first changing the order of integration _ Enter an exact form, do noe use decimal approximation _[['~&y& =...

Question

Evaluate the iterated integral by first changing the order of integration _ Enter an exact form, do noe use decimal approximation _[['~&y& =

Evaluate the iterated integral by first changing the order of integration _ Enter an exact form, do noe use decimal approximation _ [['~&y& =



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Evaluate the iterated integral by first changing the order of integration. $$\int_{0}^{2} \int_{x}^{2} 2 e^{y^{2}} d y d x$$

And this problem, you have to evaluate it created integral by first changing the order of integration. Now we're given that double integral, um 0 to 1 and another. Why one co sign X Q b ex divi now forcing it. Autograph. So in the graph, this is why I exist. This is X axis and this region here is the region off integration now double integral from 0 to 1 another. Book by toe one cosine x Q B X de vie equals two integral 0 to 1. Why consign X Q limit from bicycles to 02 Bicycles toe X square DX we got simply find the limit in tingle 0 to 1 X square consign x Q. D. X now simplifying regard Sign X cubed over three limit 0 to 1. We got the final and so as sign 1/3, there's a solution.

When I evaluate this federated into girl by converting to polar coordinates. So we need to look at what region were integrating with respects to you. Okay, So if you look at these wire regions we're integrating from, why equals zero to why equals for square root of four minus x squared. Now we can see that this is just a circle with a radius four. Okay, so here is our circle, and we're only looking at this top portion, right? This's y equals zero in this top portion here is y equal square four, minus X squared. Now, taking a look at these X cornets were only integrating from X equals zero two X equals two. Okay, so all we're looking at is this green region here. Okay, so this is our region of integration. So going ahead and taking this into account, we're integrating, right? We know that X squared plus y squared is equal to R squared. So this is really key to the negative r squared. We have our RDR D data. We always need this when we're switching to color coordinates. This is what we will later look at. And it's called the Jacoby in s. So how'd worth data's? Very well. They're going from zero to pie over, too. And our radius is going from zero to do so we can just get rid of our defeat us right away. Right. Because this does not depend on theta. So are you gonna get pie over to integrated from zero to two of our eat a negative are squared d r. And this is just interval that we do using u sub No. Okay, so we let you equal on any of our squared. Let's say do is negative to r tr s o. We're going to get multiplied by negative one half. We're going to go from zero, uh, two negative four of e to the you do you Okay, so this is negative. Hi. Over four. And when we integrate, we go from, uh, you're a teenager for so we end up with this as your answer. And if you like, you can rewrite this answer. Why? Bringing in the negative sign

So the question is, Do I drink? Are you in a way? The I printed in Kriegel, which is integration from 0 to 1. Integration from zero people express three be bye bye now moving towards the solution First, we will be waiting the inner integral zero explicit three b by which will be X y les three by limit from zero. After putting the limit, it will be west less three in tow, minus X in placido less 3 to 0. It will be quite do the West less things. So the inner integral Lewis to express six. Now we will be able with anything out very vigorous with 01 wets Does six me x It will be for immigration excess when bless six x limit from 0 to 1 one lesson 61 after putting the limit zero square less 60 which will be important. One last six that is seven. So seven will be finalized supporting the one question

Were given an iterated integral. We were asked to evaluate this integral by using polar coordinates. The integral is integral from zero to a you drove from negative square a squared minus y squared 20 of X squared y the x t y, nor to nine. With this integral in polar coordinates, I need to identify the region over which is integral. I used to find. So you see that X is going to range between negative square root of a squared minus y squared zero and that why would range between zero and a So if we draw a sketch, the exploit clean. So we have the origin and we have some A and well suppose in this case that it is going to be greater than or equal to zero course. It is equal to zero them. You simply get the integral over a line which of course, is going to be double integral over a line which is zero. And so our answer trivially be is equal to zero and we get that this integral, I will be equal to zero, which is going to be saying as a to the fifth over 15. But if a is strictly greater than zero. And we have Expo lie between negative square root of a squared minus y squared and zero. And so what this means is when y is equal to a then X will be equal to zero. So we had is sector of a circle that lies in the second quadrant. So this will be a region. Call it our and in polar coordinates. This is easily expressed as or is set of all hairs or feta such. That here was less than equal to ours less than equal to hey. And we have that it is going to live between higher too, and pie, therefore are integral When converted, the polar coordinates will be equal to the integral from high over two pi into girlfriend zero to a of x squared. Why? But now we're plugging in polar coordinates. So this becomes r squared co sine squared or sign data. And then because of the change of variable, are your key data. And we have that since the bounds on these integral are constants, we can use Beanies theorem to write this integral as a product of integral, this is the same as integral from Pi over two. Hi. Of all of our fate of functions this one, Chico Science where data time Cynthia Dictator Times the integral from zero a r squared times are attempts art or artist Fourth Dior. You know, it's simply figurative for integral earning Fourth, but we don't know one for coast and square. They didn't sign data. However, we do know if we look cosign Sprint data as Kusenov data all squared. We see that the inside of that square you take the derivative of it is found outside the square. So we should use substitution. Let you be equal to co science data and do you will be negative. Sign of feta. You think so? That signed entity data is made it to you and so are integral is now able to because of the sea. Negative. Simon won't flip the pounds on the integral so integral from you a pie. But she's going to be negative one up to you have pi over two, which is zero Oh, you swear. Do you, Heinz Integral from zero. It's a of art of the fourth D R now taking anti derivatives. Well, yes. 1/3. You cute. They read from negative 10 times 1/5 our fists values read from zero A. This is equal to when 35th is 1 15 times zero. Cute is euro minus negative. One Cute is negative ones. This is positive one and are the 5th 80th 50th state of fist and 00 seceded fifth so that we get eight assists equals over 15. This is our answer for when a is greater than zero, and it was also our answer when a is equal to zero. Likewise, if we supposed A's less than zero, then we have that are double integral, while the equal to into girl from negative, integral of negative 80 or the opposite AIDS is your own into girl from negative Square. Why squared minus B squared zero X squared? Why? Txt warn. So it's the same integral essentially as the force. Now we have a negative sign in front, and we have switched the position of a and zero, and we know that's properties of Integral. This is going to be equal to into girl from zero to negative a. The girl from negative square root of Y squared, minus squared zero expert wine t x T y. So you see, this is identical to our integral, except for now, in the first integral set for a B. So following all the same steps as before. And since this is the same as why in my way squared minus negative a squared this is really the same integral, changing acre native A. This is one to have the same answer except going exchange a native days. Now we get negative A to the fifth over 15 is equal to native. You get a 15 15. This is our answer for is less than zero. So we see that whenever the value of a we have ever integral is going to be equal to absolute value of A to the fifth over 15.


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