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Problem 1_ Use Plot3D to plot the functionf(t,y) sin {T sin(c) + y}on the region 53 < T,y < 3. Install the options Boxed->False Axes->NoneDo the same af...

Question

Problem 1_ Use Plot3D to plot the functionf(t,y) sin {T sin(c) + y}on the region 53 < T,y < 3. Install the options Boxed->False Axes->NoneDo the same after having installed the additional option Mesh->40.Problem 2. The equations+y? = 4 23 +y3 1l0 +yl0 = 1provide implicit descriptions of a couple of curves, of which the first is familiar, the others aren L. Construct superimposed representations of the solutions of those three equations, subject to the stipulation that -1.5 < T,

Problem 1_ Use Plot3D to plot the function f(t,y) sin {T sin(c) + y} on the region 53 < T,y < 3. Install the options Boxed->False Axes->None Do the same after having installed the additional option Mesh->40. Problem 2. The equations +y? = 4 23 +y3 1l0 +yl0 = 1 provide implicit descriptions of a couple of curves, of which the first is familiar, the others aren L. Construct superimposed representations of the solutions of those three equations, subject to the stipulation that -1.5 < T,y < 1.5. Do the same after having turned off the frame, turned OH pair of axes, and having installed this further option: ContourStyle-> { {Red,Thick} , {Green, Thick, Dashed} , {Blue, Thick}}



Answers

Problems $22-26$ concern Figure $16.38,$ which shows regions $R_{1}, R_{2},$ and $R_{3}$ contained in the semicircle $x^{2}+y^{2}=4$ with $y \geq 0$. (FIGURE CAN'T COPY) Evaluate $\int_{R} 12 y d A,$ where $R$ is the region formed by combining the regions $R_{1}$ and $R_{2}$.

In this problem we will cover double integral. So we want to find the integral of this area S which is comprised of the region's R two and R. Three of the function X. And so I have written here in green are double integral. And to make things easier, I made the inside integral pertains X. While the outside integral pertains to Y. And we see that are bound for X are going to be negative one. And of course the square root of four minus y squared, which is just this rewritten to solve for X. And ry bounds are going to be from zero two and without the way we can proceed to solve this. So keeping our outside integral the same, we have that the anti derivative of X is going to be X squared over two. This is from square to four minus y squared. It's a negative one. Why? And so we get one half times four minus y squared minus one half he like and this becomes two minus a half Y squared minus one half again. Dy So the integral of 0-2 of 3/2 minus one half Y squared. Do you know why? And taking the anti derivative of that perspective way, we will get 3/2. Why minus 1/6. Why huge from 20. So we have to do is plug in two for what? So we get three square in terms too minus 16 times two cube. This is going to be three minus four thirds. And that leads us with A value of 5/3

In this problem we will cover double integral. So for our first double integral we have that. The order of integration is first with Y and then X. So are inside integral has the bounds zero and negative square root three times X. And these corresponds to the white bounds So these and are outside integral has the bounds for X. So that that's going to be negative one and zero. And since we have that out the way we can proceed to solving this, so are outside integral stays the same and the anti derivative of two. Why in this case is going to be y squared And that's from negative square over the three X zero. Outside it's DX This leaves us with the integral from negative 1-0 of three, X squared dx in the anti derivative of that respect to X is going to be X cube. So we have x cubed 0 to -1 And we get zero minus native one. And that's just going to leave us with positive one. Moving on to our second double integral. We changed the order of integration. So now we do X first and then why? So are inside integral has bound relating to X. So we have a negative one and negative Y over the square root of free. Which is this. If we solve for X and are outside integral has the bounds relating to why. So from white zero. So y equals the square root of three. And so we have that out the way. So we can solve this are outside integral stays the same and we take the anti derivative of the inside with respect to X. So that gives us two X. Why? And we want it from negative y over the square root of free two, negative one. And that gives us we'll see negative two. Why squared Over the square root of three minus or plus? It's plus too. Right. And this outside should be. Do you know why we can break down this interval when I first splitting it and then putting the constants in front of the interval so we can get negative two over the square root of three Times The integral from zero square 23 times are not times of y squared, do you? I plus two. So I'm the integral from zero square in a freak of why do you why? And this gets us -2 over the square root of three times Y cube over three from Square 123 zero plus two times. Why squared over two From Square 123 20. And this should yield us an answer again of one

In this problem we will cover double integral. So for this problem in order to solve it, we want to Use two types of double inter rules. One using rectangular coordinates and one using polar coordinates. And I have written in green the devil integral. Using regular rectangular coordinates. So it deals with X and Y. And we see that the inside integral pertains to values of why. So we have that are lower bound zero and that our upper bound is square root of four minus X squared. And that corresponds to the line Y equals zero bounding. Our region are one below. And these semicircle bounding it above and the outer integral pertains to the value of X. And we see that our region Goes from -1 -1 -2 on the x. axis. And now we perceive to solve it. So are outside integral stays the same. And our inside we get y squared square root of four minus x squared zero X. This becomes our interval mint. 2 -1 of four minus x squared D X. And taking the answer derivative of vet. We get four x minus one third X cubed From two and 1 plugging in negative one first we get negative four mm plus one third because negative one cube just negative one. And those two negatives make that one third positive minus negative. Eight plus eight birds. This simplifies to For -7/3 Which gives us a final answer of 5/3. So that is our answer for our first double integral with rectangular coordinates. And now we move on to our double integral with polar coordinates. So in black I have that double integral and we see that the outside integral pertains to the value of the angle theta. And it goes from an angle according to the graph uh two pi thirds and hi. And the inside integral which pertains to the value of r the radius should go from 1 to 2. And we change our function to why into to our sign data. And because D. A. Becomes R D R D. Theta we just multiplied. This are with the one that was meant to substitute why? And we get R squared. So now we solve that. So are outside integral stays the same. And on the inside for anti derivative since we're only looking at the our term we can move to sign theater to the outside and within our parentheses we get Are huge over three from 2 to 1. And that gives us a clear outside into Rome the same. Hi. So we will get eight thirds minus one third Which just becomes 7/3. So we get 14 thirds Sine theta D theta to make things easier. We just moved that 14/3 to the outside of our interval. So now we're just dealing with this integral signed data do you say to And the anti derivative of sine theta is going to be negative coastline data. So we have this and this is from hi to buy thirds Here we had 1403 times joseph I is negative one and that extra negative just makes it positive one. And we want to subtract again. Coast of two pi thirds is going to be one half Or negative 1/2. And since we have the negative here, that would make it positive one half. So we have yeah. And that inside of parentheses is going to be one half and one half times 14/3 is going to be in a simple right version. 7/3.

In this problem, we will cover double integral. So the first way we can write our double integral for this problem is with the order of integration as being why on the inside and then X. On the outside so we have here that are bound for the inside integral pertaining to why are going to be zero. And the square root of four minus X squared which corresponds to the semicircle here. And that the bounds for our outside into rope pertaining to X are going to be negative 2, 2 negative points. So we proceed to solve this and keeping the outside integral the same. We know that the anti derivative respects why on the inside is going to be y squared and it's from the square root of four minus X squared zero D. X. And that leaves us with the integral from negative to negative one of four minus x squared T. X. And taking the anti derivative with respect to X. The inside we get four x -X cubed over three from -2, -1. And this becomes negative four minus negative one, third minus a negative eight minus and negative eight thirds, which ultimately Leads us to a value of 5/3. Now our second integral. We reversed the order of integration. So now are inside integral pertains to X. While the outside integral pertains to why. And we see that are bound for the inside integral becomes negative one. And the square root of four minus y squared which is the semicircle and our bounds for the outside integral chain, why R zero. And the square root of three which is here. So now we proceed to solve it. So we keep our outside integral the same. And the anti derivative of the inside with respect to X is going to be to Y X. From the square root of four minus Y squared zero. So this becomes the square root of zero and not the square zero. The integral from zero to the square root of three. Uh Two Why square root of floor minus Y squared? Sorry, this was not supposed to be zero. This was supposed to be -1. So we get plus two white an MDX here. So now I'm taking me anti derivative with respect two x of the inside we should get are not respect to X. With respect to y. I don't know why I'm blundering to me. Should be working straight away. So we should get negative one third four minus Y squared To the three heads plus. Why squared? This is from the square root of 3-0. And you plug in the square root of three And we get negative 1 3rd china's one was three. And we know that three minus a third is going to give us five thirds again


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