Question
Let T be the triangular region bounded by the lines x + y = 1,x = 0, and y = 0. Evaluate the Iine integral If coS X+y) dxdy:
Let T be the triangular region bounded by the lines x + y = 1,x = 0, and y = 0. Evaluate the Iine integral If coS X+y) dxdy:
Answers
Evaluate the double integral. $$ \begin{array}{l}{\iint_{R} x \cos y d A ; R \text { is the triangular region bounded by the }} \\ {\text { lines } y=x, y=0, \text { and } x=\pi}\end{array} $$
Hi have you given white refined world sex trivia From Mexico to zero. Let's go to do with the Y. Axis wise. Plus posted by over two narrative over to the ST uh provide data between this sign Will sex and uh Y. Accident is there we go. Here we have. Let's go by we just go by here we have 02 zero to pi over two. This is by over two. This is zero to pi over to the ex we take from here X. DY this area and it's symmetrical with respect to origins of just two times that we have so you get the required India. So this will be given us two times they had to pay over two two times 02 By over two x. g. one. So two times there are two by over to access sign wine. And even so all the studies coming out to be to the violence. So thank you.
We want to find the area of the region that is bounded between these curves. So here you see the graph of the three functions and here is that region bounded between the curves. Uh So we want to find this area and this is going to be a straightforward definite integral because you can see this purple function X plus one is staying above uh The black function cosign X. So the area of this region is simply going to be The definite integral from X0 Up to X. S one of the function on top X plus one minus two function underneath. Cosine X. So integral from 0 to 1. in a rule from 0 to 1 area Will equal the definite integral from 0 to 1 of the function on top. Purple that was X plus one -2 function underneath which was co Synnex deok So uh this definite integral will be the area founded between these curves. No uh the anti derivative of X is X squared over two Plus the anti derivative of one which is X. The anti derivative of negative code syntax is negative Synnex. I take the derivative of uh Synnex, I get co Synnex. So derivative of negative cynics will be negative cosine X. Okay, so this is the anti derivative of our integrated X plus one minus cosine X. We need to evaluate this between X zero And X is one. When next is one we have one square which is 1/2. So we have one half plus one ah minus The design of one. And then we need to subtract this same expression evaluated when X zero when X zero, zero squared over 20, X itself is zero and then the sine of zero is zero. So we're really subtracting zero. Just double checking integrating between X zero and X is one. So when X is one, we do have to take the sign of what? Okay, so it was just kind of weird but that's what it is. All right. Um So one half plus one minus sine of one. Now sign of one, make sure that when you use your calculator uh that you calculate set on radiance because that's really the sign of one radiance. A sign of one Comes out to be .841. Uh at .5, that's the one half had one. Yeah. My mistake, I'm supposed to track to sign a 1.5, that's the one half plus one. And then subtract, sign up one And we get .6585. So .659 2 0.659 is our area. So this area and here is going to be .659.
Okay, problem 10. That's, uh, graft these lines. Michael's two straight. Why equals X will be like this and the interesting about this when X equals zero. Likely wants to this point here, this is why equals one offering in this equation here when X equals zero while also equal. Once we know that this equation this line here has passed through White was one so look like that. So this is the battery. We're trying to find the area into the area. Um, going up in a role. Well, we're gonna have this line here. Br lower boundary and this exponential curve your upper boundary. So the one goal on the X to do y so we saw you dio rewrite this equation and its equation in terms of text. So for a lower boundary between just equals one minus y in our boundary Natural log. Why 12? So, 12 Uh, that's a lot of why miners One Why do you want? And this is X Y national audible. I'm last wine minus. Why close? Why? Where it over to? This is why that's what Why minds to oy life's good. A two on 12 have to natural lot of to minus for for two minor. So we turned one here with a zero from zero minus two 1/2 services to Natural Lago, two miles for plus two, plus two minus 1/2 to camp with zero in our final answers to natural log of two minus 1/2.
It's been problem 62 were asked to find the region bounded by the curves sign of X and the Ark Sign of X when X goes from 0 to 1/2. Easiest way to look at this as graphically. So if I look at a graph of this scenario So I've graft to things here I've graphed the sign of X So the sign of X you see in red and then the arc sine of X you see in blue And I'm looking at the interval from 0 to 1 half looking and so it looks like the arc sine of X is larger. So that's the curve, this on top. So to find the area between the two curves, I would just need to be able to find the integral from zero toe 1/2 of the ark sign of X minus the sign of X de eggs. So that's what I would need in order to find the answer to this particular question. Um, so let's go do that. So, um, so in this case, my solution is going to be the interval from integral from 0 to 1/2 Thea arc sine of X minus the sign of X dx, and we've done this several times. This section with integration by parts. That's how you handle this arc sine. And when you do that, you're an end up with the enter one minus x squared the arc sine of X Excuse me, um, one minus X squared and then plus X, the arc sine of X. And then when you integrate the minus sign function, that's where you're going to get the co sign function and all of this evaluated from zero to 1/2. And so now we just substitute these values in. So in doing that, So if I substitute 1/2 so one minus 1/2 squared So that's one minus 1/4 that's 3/4. So this is the square to 3/2 is the first term, plus 1/2 the arc sine of 1/2. What angle gives you Ah sai value of 1/2 That's pi over six plus and then the co sign of 1/2. Now substitute zero and you're going to get the square root of one and then plus zero. And in the coastline of zero is also one. So what? We end up with here is the square root of 3/2 plus pi over 12 plus the co sign of 1/2 minus two. And then if I substitute into my calculator, this is about 20.5407 And that should match what I got on the graphs. And it s so if you look at what I have here on my grand 0.5407 I got it by substituting the actually doing the new miracle integration. And by seeing here is the value of what I just obtained a moment ago in this particular answer, so