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A) Let f(T,y; 2) = 212 +3y2, and let p and q be the points p = (1,0,0) and q = (2,-1,1)_ Use the formula from I(c) to find Jc df , where C is any curve that starts ...

Question

A) Let f(T,y; 2) = 212 +3y2, and let p and q be the points p = (1,0,0) and q = (2,-1,1)_ Use the formula from I(c) to find Jc df , where C is any curve that starts at p and ends at q. b) In the previous written homework; you found f -ydr + xdy over the unit circle in the zy plane, and should have found the answer 2T . Why does this fact mean that there is no function f(r,y) such that df ~ydr + x dy? c) Use the fundamental theorem of line integrals to compute Jcry? d +xydy, where C is the curve p

a) Let f(T,y; 2) = 212 +3y2, and let p and q be the points p = (1,0,0) and q = (2,-1,1)_ Use the formula from I(c) to find Jc df , where C is any curve that starts at p and ends at q. b) In the previous written homework; you found f -ydr + xdy over the unit circle in the zy plane, and should have found the answer 2T . Why does this fact mean that there is no function f(r,y) such that df ~ydr + x dy? c) Use the fundamental theorem of line integrals to compute Jcry? d +xydy, where C is the curve parametrized by r(t) = (cos(3t),1 + sin? (t)) for 0 < t < T.



Answers

Let $C_{1}$ be the curve $x=\cos \theta, y=\sin \theta, 0 \leq \theta \leq 4 \pi,$ and let $C_{2}$ be the curve $x=\cos t^{2}, y=\sin t^{2}, 0 \leq t \leq 2 \sqrt{\pi}$. For $P(x, y)=x^{3} y$ and $Q(x, y)=y^{2} x,$ show that $\int_{C_{1}}(P d x+Q d y)=\int_{C_{2}}(P d x+Q d y) .$ Explain why this is so.

Are prompted to three Given our of tea is a co sign T B sine T and C scientist e probably show sees a circle centre of the origin prided by a squared equals B squared C squared. So given our first find the magnitude of our of tea which is going to be the square root of a square co sign square T plus B plus C square times sine squared t and that equals squared of a square CO sign square T plus a square signs square T which is you go to a Therefore, our tea is good. A co sign t was a comes Jericho zero plus sine of t zero comma Beacom Ah see? And therefore you can see that see is a circle centre at the origin with radius a line the plane of these two vectors. All right, so find the are calling for the circle C, which is just to go to in agro 0 to 2 pi of our promised t my interior DT ingrown 0 to 2 pi ah square of a squared sine square T plus B squared plus C square times co sign square T d t. That's you go to the integral from 0 to 2 pi of a square times sine square T plus co sign squares T This is a squared this whole and grow becomes integral from 0 to 2. Pi of a, which is just two pi times a a part c So, Marcy What's assumed that the curve is a co sign t plus b Ko Sai Inti Don't be sign of T c ko 70 plus de sine of t and e co sign t plus f Sigh Inti. So find which conditions that this curve describes a circle and find his our claim. Okay, so are of tea magnitude. It's gonna equal each one of these elements squared at it together and then working a group The co sign squares together signs core together and the sign two teas together So we end up with a squared plus C squared plus e squared times coastline square T buzz Be square plus d squared plus F squared sine square T plus a B plus C depot CF of sign to T DT. All right, so if a score of pussy Scorpy e squared equals be Square Pussy's Corpuz F squared and a B plus C depots ET is zero. Then we get our of tea to go to a squared plus C squared plus C squared, which represents ah circle so that conditions would be a square pacy square. Pacy square equals B squared plus D squared plus F squared and a B plus C D plus e f has equal zero. So now defying the arc length, take the integral from 0 to 2 pi of the magnitude of our vector be co sign team. I saw a sign T deco sighing t on a cease i Inti and f Co sign T mice. E sai Inti DT becomes integral from two pi of screwed of a squared plus C squared plus e squared signs. Corti Cosby's Corpus D Squared plus F squared of co signs Quartey minus a B plus e d plus cf assigned to t. And since we said thes two are equal and this is the go to zero. Then we get into girl from 0 to 2. Pi integral a square pussy's Corpuz East Grayer D T, which is the Goto Integral. A score plus, he scored plus C squared times t that air from 0 to 2 pi or two pi times skirt a squared plus C squared plus C squared

Okay, This question wants is defined the line integral over these two straight segments. And we already showed that this vector field f is conservative, so we're only going to have to look at the start in the end. But to use these points, we're going to have to find the potential function for F. So since office conservative, we know that f is the Grady int of some potential. So in order to find the potential, we just take respective anti derivatives of each component. So our first contribution comes from the anti derivative of the axe. And that's just the X anti derivative of two x y squared DX. What should be X squared? Why squared, plus some unknown function of why that may or may not show up. Then if I, too, would be the integral of to X squared why plus two. Why de y giving us another x squared y squared then plus a y squared with a possible function of X and we see that this function of why is accounted for. So this is our function correct up to a constant. So fyi of X y would then just be X squared. Why squared plus y squared. So the integral oversea to of f dot de are. Then I would just be five The end minus fly of the start and the end is at 31 and the stark is out 11 and 531 is nine plus one. So 10 and five of 11 is one minus one. So our line in a girl just has value 10.

In order for us to evaluate the surface integral of Stokes, assume just the double integral along the surface of Dell Cross f dotted with the unit normal vector D s. We first must select a viable surface on which to evaluate this. And because we're given in the problem that the boundary coop is X squared plus y squared equals 12. If we select Oh, surface based off of this was Sofus is going to be 12 minus x KWood minus y swallowed. It makes the evaluation of this into role fairly easy. So then all we have to do is start evaluating this interval. So starting with this Del Cross F if we set up a matrix with I J K. And then it's Dell first. So that's the postal with respect to X with respect. A Why, with respect to Z and then which is two why negative z and X. This is equal toe one negative one and negative too, which is found by taking the discriminative of this matrix. So now that we have Dele Cross F, we have to find the unit normal vector, and this is going to be equal to the ingredient of always surface. You calling s divided by the magnitude because it's a unit. A normal vector. The ingredient of a surface here is going to be negative. Two x for the i component negative to why, with the J component and one sweeper K component divided by the magnitude here which is going to be the square root of four x squared plus four y squared plus one. Now, if I frank this so we have more space to maneuver Now we have two dot ah unit normal vector. Just you, with, uh, still cross us. So this leaves us though cross us dotted with the unit Normal vector is going to be equal to weaken. See, it's one times negative two x so that negative two x and then negative one times negative to wise That's plus two Why and negative two times one which is negative two all divided by squirt of four X squared plus for y squared plus one. And before we plug this into the integral, we have ah, Dell cross f dotted with the unit Normal vector in terms of accent. Why over this is a DS and in order for this ds to be in terms of X and y we have to If we say this is ah, central function D s is going to be equal to se. The potion will suspect the X Schoon plus the postal Inspector y squared plus one d A, which is the X t y. So if we take this quote of thes poche ALS and one we see that DS is going to be equal to swell loot of four ex quote, which is the postal with respect to X. It s squared. Plus for why screwed close one, Do you? A. So now if we set up our interval, you see this integral? Well, a leader now is going to be equal to negative two X plus two y minus two All over this quote off four X squared Plus for why squared plus one kinds the square root of four X squared plus four y squared plus one e A. And as you can see the squalid, you will cancel each other out. So if we again shrink this down to get more room to manoeuvre, all we have to do now is evaluate this integral, which I will we wait help you. So we have this double interval of negative two X plus two y minus two d A. And however this area of which integrating is going to be this circle X squared plus y squared is equal to 12. And because we are integrating basically the area of this so call here if we convert toe polar coordinates meaning we put X and y in terms of all in Seda, it will make the limits of these intervals much easier. So if we say X is equal to ah coastline Fada, why is equal toe all times? Sign of Fada and D A, which is equal to D Y D X is going to be equal toe Ah de are de so then if we plug this in, if we first look at a balance weaken, see that all is going to go from zero to the squad of 12 and Seda will be going from 0 to 2 pot. So we have ah, double integral off Negative, too Times X, which is Ah, coastline Seita plus two times why, which is ah sine theta minus two, all multiplied by all D. O D. Seda. And of course, the limits of intervals here, uh, going to be all just from zero toe screwed of 12. And then they'd off with this out of one, which is from 0 to 2 pi. And again. Let me shrink this. So we have more room. Two minutes. It appears oath Ada got caught, but it will related. So now if we evaluate this integral food with, uh you see, it's from Monsieur two pi, and if we distribute this all, you get negative to all squared coast data plus two squared sine theta minus two R D o d theta. Which is of course, equals from 0 to 2 pi of negative 2/3 uh, cubed whose data? Plus 2/3 a cube sine theta minus our squad from zero to the sward of 12 di fada. And again it oppose. We're running out of space. So if we shrink, move this up top. And if we plug in our screwed of 12 0 we're left with the integral from zero. Your two pi of negative 2/3 times 12 12. Coast data plus 2/3 time is 12 Moot. 12 Sign Fada minus 12 de fate up. So now again If we integrate, we have negative 2/3 times 12 boot 12 sine theta because we've integrated plus 2/3 times 12 Route 12 Negative co sign data minus 12. Data from 0 to 2 pi. And so if we look at this first Pote because sign of two pi and sign of Zago of both zero, we can disregard it. And if we look at this second tomb who co sign of two pi is one as his co signing zero. That means we're basically subtracting, uh, the same thing from itself. So that will also go. It is, you know, So this just leaves us with 12 data 0 to 2 pi which is equal to negative 24 pi.

Were given a vector field Big F and A Curve C where, as defined in part a function little f such the big F is equal to the ingredients of little F. So big f is the function y Z I plus x zj plus X y plus two zk and C is the line segment from 10 negative too. 2463 Now in part A. We know right away that the partial derivative of Little Left with respect to X is equal to the X component of big F, which is Y Z. And this implies that little F of X y Z is equal to x times y times z plus some function G of y and Z, and therefore it follows that the partial derivative of little Left with respect toe why this is equal to X Z, plus the partial derivative of G. With respect to why. But we also know that this should be equal to the why components of big F, which is xz. Therefore we have that the partial derivative of G, with respect to why is equal to zero, and this implies that function G is equal to some function h of z to take the anti derivative with respect toe. Why therefore something this together we have that little less is equal to X y Z plus h NC and taking the derivative with respect Dizzy little left with respect, dizzy is equal to X y plus h prime of Z. This is also equal to the Z component of big F, which is X y plus two Z. Therefore, it follows that H prime of Z is equal to two Z or in other words, HFC is equal to Z squared, plus some constant K. Now, if we take a to be equal to zero, we have that little left is equal to X y Z plus Z square in part B, whereas to evaluate the line integral of big F along the curve C by the fundamental the're, um for line into girls. The line integral of big F along the curve C is equal to little less at the terminal in 0.463 minus little f at the initial endpoint 10 negative, too. Plugging these into our above formula For little left, we get 81 minus four, which is 77


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