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Find each double integral over the rectangular region $R$ with the given boundaries.

$$

\iint_{R} \frac{y}{\sqrt{2 x+5 y^{2}}} d x d y ; \quad 0 \leq x \leq 2,1 \leq y \leq 3

$$

$\approx 1.6874$

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Baylor University

University of Michigan - Ann Arbor

Idaho State University

Boston College

All right. So we're giving the integral double Integral over region are of the function y over this square root of two X plus five y squared d A. And then our region are is basically the set of all points such that zero or Xs between zero and sue and wise between one and three. So just reading this out again, this is a little set notation here basically saying that this region are this set is equal to all the points x and y such that this colon here x is between zero and two And why is between one and three? Okay, so just drawing out region are should look a lot like a rectangle. So just straight line, baby. Okay, so 0 to 2. And then why bring just from 1 to 3? So we have one. 35 equals three y equals one X equals zero X equals two. So it should just be this region right here. It's actually a square region here. So now we have to set up the double integral here, so nor to write the correct limits and the correct places. Um, what I'm going to do is evaluate this enroll. Respect to why? First. Yet It's a lot easier to do wipers and you'll see why I went on when I started evaluating this. Why choose such respect? The wipers of our functions. Two X plus five y squared. Um, okay, so now when I was learning this, it always tripped me up. About what limits? Go where? And basically the tactic that I use now is I remember that you always evaluate these intervals inside out. So we evaluated, spent two wipers and then respect to X. And then when we're anyone who spent two wipers we want to think about how does why range? Basically for a region are how does wide change or range? So why ranges from 1 to 3. Sarah, let's go from 1 to 3, and then X ranges from 0 to 2. Okay. And this would be our double integral Here. Um, now, evaluating this could be a little tricky on May not see it immediately, but I will show you what to dio. Do you? What? So we're actually going to use, um you substitution here and we're gonna let u equal this inside right here because we're always looking for a derivative and its anti drip or it's the A function and it's derivative. Okay, So you'll I'll show you what I mean by that in a moment. So because we're in a race, but to why here? We should be driving respect. Why here? So we have to you is evil to, uh, 10 wife. And then Plus, since X is a constant in this case because we're drying spent two why, there should just be zero. And then this is times do you? Why? Okay, so we have do you equal to 10? Why d y? And then I'm just gonna move the tens of other sites. So do you. Over 10. So we have. Do you over 10 is equal toe Weide. Why? And you can see that basically, this is our function. And this is its derivative, in essence. Like we can get, um, this by deriving if we correctly use U substitution. Okay, so plugging everything in we need to interchange are X or R y stuff with are now new limit of you. So we have widely wise eagle to do you over 10. So we have the 1/10 out here and then do you just taking this constant out first? So now we have one over the square root of you because you're allowing this to be you in this case. And then look at the limits. Basically, we want to plug our white limits to get you limits on this right here. So plug in one for why we get five plus two X and then we have when we plugged in three. We get 45 plus two x. So now we can evaluate this inner girl. This should be equal to 1 10 times. Uh, actually, I'm gonna rewrite this as basically you to the neck of 1/2 first. And then we can use our rivers parables. So we add one to the exponent 1/2 and then bye bye. That new experiment, So 1/2. This should be evaluated from five plus two x 2 45 plus two x. And don't worry about limits now, having variables in that you're gonna encounter. There's a lot. But basically, if this limit out here and here is the same as our outermost integral, then it will be basically transformed into a numerical answer. And then because we're gonna be basically integrating respect to this variable and then so it's always gonna result in a numerical answer. So don't worry about limits being variables in this case. So we have 1/10 times two times you to the 1/2 of this and just playing you done. We have one time tires two times two plus two X plus 45 to 1/2 and then minus what we get when we plug in the bottom limit two times two x plus 5 to 1/2 power that we want to put this back into our inner inner girl because it's a double integral. So we have one time. Has a little from Syria to was taking this, um, constant out. So we have a role Absurd. Teoh two times two X plus 45 1/2 to 2 times two x plus five 1/2 and you know the drill. We factor out this fat, too, because it's a constant and we pull it to the outside. So we have 1/5 times the integral from 0 to 2 of two x plus 45. It's the 1/2 power minus two x plus 5 1/2 power DX. All right, now to evaluate this, we think, um, basically, since we have a linear functions inside of our power function, you want to think about these as if they were just, like, ex, the 1/2 and how you integrate excellent half. We just use the reverse spiral. And it would result in this basically I'm plus C, but we don't really care about see, because it's the definitely rule. So we're gonna integrate these as if they were just a regular axe. But we're gonna do a little trick right before, um or right after doing that. So this stuff we helps with cutting down on the amount of time you take to solve these because you couldn't use u substitution here, um, to evaluate these. But this linear trick is a lot easier and actually comes from you substitution. But it's just a little shortcut that I used. So you have that 1 50 times. Basically two x plus 45. Just treat the inside as if it were a regular X. So we have three halves, 2/3 because of vision by three houses, same as multiplication by 2/3. Um, we divide this all by the inner linear coefficient. So not be too and then minus. And that's a little trickier. So remember to always abide by that linear coefficient after you do the irregular. And it was with a lot of our all other functions that you can anti drive. Basically, if you have a linear function inside of another function, you can take the tiger over the outside function and then divide by the linear coefficient on the inside. It always works. Um, for a 2nd 1 we have same thing 2/3 times two x plus 5 to 3 halves. Then this is all divided by two. Because our linear coefficient here is to so basically, take the anti derivative as if it were regular X and then divide by the linear coefficient at the end. Okay. Oh, and remember to evaluate this from 0 to 2. All right, so now we're just going to take out all of our constants really quick. So we have, um basically, after you pulled all these out, it should be a 1/3 in the end that you pull out because, um, you have 2/3 to buy by two, which is just one Thursday. Just play out the 1/3. So we have 1 15 has 1/5 times. 1/3 is 1 15 that we have two X plus 45 the three halves minus 20 no, to s plus five to the three halves. Body weight from Syria to All right. Now, plug in this in plugging in to first our top element, we get four plus 45 three house minus nine, 23 house and my ass. What we get when we plug ins room So we have 45 to 3 halves and then 5 to 3 halves should be able to one fifth and we're remembered. Suits attract the entire thing. So always put parentheses around the value you get when you plug in the limit, because it just helps with, um, just about doing this and making sure that is correct. So we have 1/5 um, times 49 3 halves, minus 93 halves. And then we have minus 45 to 3 halves and then plus 5 to 3 halves because we're distributing this negative to all the components in the inside, said because 45 is positive. We make that negative. And since this is not going to make that positive, and that's why it's important to put the parentheses because we remember to distribute it to the entire quantity. So they have equals one times the square root of 49 to the third minus squared of nine to the third, minus 45 3 halves plus 5 to 3 halves. There should be equal to 1/5 times seven to the third, minus three to the third, minus 45 to 3 halves plus 5 to 3 House. And this should be able to 10 did I do this room? Oh, my God. I forgot the sorry. Sorry. Sorry Meant 1/15 every time I forgot the one. Uh, yeah. Okay, so we have 1 15 times. Awesome to third is 3 43 and then three to the third is 27 that we just keep the other two the same because simplifying them would result in a decimal on. This would be our final answer.

Rutgers, The State University of New Jersey