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# Integrate $f$ over the given curve.$f(x, y)=x+y, \quad C: \quad x^{2}+y^{2}=4$ in the first quadrant from (2,0) to (0,2)

## 8

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Okay, folks. So we're gonna be doing problem number 29 here. Um, so for this problem we have, we have a circle, and we're gonna be integrating a function f along, um, along the circle in the first quadrant from 20 to 0 to so before before we get started doing the integral Let's let's just get a few preliminary steps out of the way. So for a circle, you have this equation. Obviously, you have expert plus y squared equals four. And from this, you can solve for why, and that is going to be route route four, minus x squared. Um, what, you know, because we're integrating along the first quadrant. So why is always gonna be positive and why prime, which is which is a derivative with respect of the variable X, is going to be negative x over root for a minus, X squared. So that's why prime, and from this you can get DS and DS is gonna be a positive number, because that's because DS is a is the length of, like, a really, really small triangle which can be written as the X squared plus y squared. So this little length here is what we call it the s. Excuse me. My in writing is really horrible. The S. So this is the X, and this is D Y. So the S is always going to be a positive number because it's a It's the length of failure of the of this side. I forgot what this side is called. Uh, eso es is always going to posit number because it's the length, okay, and you cannot have a negative. So now that we're clear on the fact that the U. S is gonna be a positive number, we're going to Ah, plugging in the, um we're gonna do is we're gonna divide out the ex inside the square root and multiply it by back in on the outside. Okay, so we have one plus d y over the X squared, but do you buy over? DX is really just white prime. So we have white prime squared. So we have, um, X squared over four minus x squared, and this could be simplified into two over square root of four. Mine is X squared DX. Okay, So this is our DS, except, um, except when we're actually integrating the function along the curve. When we're integrating the function along this curve, the D X they were considering is actually negative. Okay, Because because this is the direction they were integrating over. And we're going when you're going over the curve in this direction, your DX, as you realize it's in, isn't this direction okay? The excess negative. But you don't want that to happen, because when you have a negative DX, your D s is Your DS is going to be negative as well, because this number right here, this factor is always gonna be positive. And when you have a negative DX, that means the SS negative. And you don't want that to happen because the SS and cannot be negative. So the way we're going to do this is we're going toe. Just insert for this problem, insert a factor of negative one just so that we don't end up with a negative DS, which is unfit sickle. Okay, so now we have all of our preliminary steps. We can just plug in everything. Um, we have f if x Why ds and from 2 to 0 x plus Route four minus X squared. That's for what that's for a That's for the function f um multiplied by D S s o. D s is going to be too over route four minus X squared, multiplied by negative DX. And now I'm gonna split this into two parts. We have zero the to two x the X over square root plus twice, 0 to 2 the X And for the first, integral. If you do ah, you substitution, Which is, um, which is a technique where you define a new variable as U equals four minus X squared. And when you define when you define this new variable, you can get D u is negative two x x and you see here you have two x t x so that you can plug it. You can plant that. You can substitute that with negative, do you and you crank out the algebra. You're going to end up with four for the first integral plus two times two because I think this is pretty obvious. So we have four plus four, which is eight. And that's the answer for this problem. We have eight. Um, thank you very much for watching, and I will see you next time

University of California, Berkeley

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