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Problem 27

In Exercises $27-30,$ integrate $f$ over the give…

Problem 26

Evaluate $\int_{C} \frac{1}{x^{2}+y^{2}+1} d s$ where $C$ is given in the accompanying figure.


$$\int_{C} \frac{d s}{x^{2}+y^{2}+1}=\pi$$


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Video Transcript

you guys. So today we're gonna be working on a line integral. So we have the integral of f of X y oversee using X Y. Because there's two components is it's a two dimensional shape. In our case, f of X y equals one over X squared plus y squared plus one and testify. And that's going to that oversee, See, is given here in this graph, which is pretty simple business. You're so square in our case, lucky, lucky us. So we're actually gonna do find divide this graph into four different parts. You see one C two, C three and C floor. So starting with C one, we're gonna write in terms of our tea, we've got our of tea equals. So we're going out with her ex, which is right here. That's what you got One tea times. I plus moving on to our why, which is zero zero t vector j. So from here, we know that t is defined on the interval from zero Sorry, positive one. And we can also find the magnitude which is gonna be the length of the velocity vector. So when the velocity vectors square root of X squared plus y squared exports and meet one squared and wise words to me zero such meat one. Now we got C two r of t equals In this case, the ex isn't moving Tze with zero t i plus one So one t j also defined from zero one and velocity is at the los e vectors actually be the same You see the square root of X squared and why square DHS is your run one So it's one squared equals one c three We've got our of tea equals the X is moving in the opposite direction. Now we've got negative one t vector I plus wise not changing so zero y zero t sector j again we know it's defined t's greater than or equal to zero and less than an ankle toe one and length of a lost the vector Square X squared which is negative. One squared which seems one squared zero squared So one last one c four The art of tea cools moving downwards So negative one No, Sorry X is not changing about zero zero for X t Director. I plus negative one. Why so negative? One T Specter J again to find I'm t between is greater than or equal to zero and less than one last lost e vector square root X squared And why squared? She ends up being one squared, which we know was one awesome Berries. Now we can move on to our equation. So based off of the rules of my into roles, we know that the integral of C. Beth Lex Why his same as adding up all the individual parts. So he's got in a roll. C one, uh f x y d e s plus the integral see too of X Y yes. What's in a girl? Three effort X. Why D s and in a row c four x y Yes, based on the work, all the work we did here we have all we need to plug into this equation. So we know. So for sea ones to find from 0 to 1 and our f sovereign plug in our X and our why one tea is just tea and zero t zero plus the magnitude was one plus super C +201 of x and y against zero and teeth one. Now we want to see three from 0 to 1 exes. And why so negative? T zero 9 to 1. Duty finally for C 4021 zero Negative. T magnitude one. We know our equation. We already to find up here so we can actually plugged into that for our next step. Yeah, groups are f of X is up here. So it was me one over x squared t squared plus y squared zero squared plus one again from INA grow 01 one over X squared to zero squared plus y squared t squared plus one in the girl again from 0 to 1 one over X squared should be negative T Square, which is the same as T squared. Wai scored a zero plus one U T. Again and the girlfriend 0 to 1 one over x squared. It's zero squared plus t sport. Why square was t squared plus one Dean T. Since it is a little repetitive, we can simplify it down by saying it's four times being a girl from 0 to 1 of one over t squared, plus one DT with These are integral rules. We know that the in a grove of one over t squared plus one is gonna give us Arc Tanne. So inverse tangent of tea. And we know that we're going from 1 to 0 when you can't forget this four. Put it out here. So now we're gonna plug in our top number and our bottom number. Subtract four again. Inverse tan one minus inverse tan of zero, which was four times pi over four. The name is our grand finale. Answer. Hi.