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Evaluate $\int _ { C } ( x + \sqrt { y } ) d s$ where $C$ is given in the accompanying figure.

$\frac{1}{6}$$\left(5^{3 / 2}-1+7\sqrt{2}\right)$

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Campbell University

Oregon State University

Harvey Mudd College

University of Nottingham

Okay. What we want to step through is we want to evaluate, um, the integral over the curve of X plus the square root of Why D s, um, where see is actually going to be, um, is equal to the union. Uh, the curve, C one and C two. And it is given in, um, it is going to be given in this graph right here. So on the X Y playing, and so we have, um, portion of this curve C one somewhere Why is equal to x squared? And then, um, we also have, um, then down here, Um, where, um we're gonna let see to be Ah, Why equal to X. And, of course, this point up here is 11 Okay. And this is 00 Okay, so now we're going to let um, um, see one equal to that curved path. And so that is gonna be our if represented by r of T is equal to, um, t i plus, um t squared j um and therefore um and then see, too is going to be represented by r of T is equal to t I place t j um and then of course, we're going from tea for both cases. T is going to go from 0 to 1 inclusive. Okay. And so, um, now we know that we've got to find V of tea, which is represented by the derivative of Are. So this isn't B I plus to TJ. So the magnitude of V of tea for the first curve is equal to the square root of one squared plus two tea squared, which gives me the square root of one plus four t squared. And so, for the first curve, D s is equal to that square root of one plus 40 square DT. Okay. And then for the second curve, uh, B of T is equal to that derivative, which is gonna be I plus J. And so the magnitude of e of t is gonna be the square root of one squared plus one squared, which is gonna give me the square root of two. And so D s is the square root of two TT. Okay, so now what will you need to do is to go ahead and set up our inner girls. And so the first integral is going to be from 0 to 1. And it will be, um, X, which is t plus the square root of why, which is the square root of T squared? Um, times, lips, um, times, um, the integral, sometimes the square root of one plus four t squared tt. That's so That's the first integral plus the 2nd 1 from zero toe one of in this case, um, we have, um t which is X plus the square root of why, which is a square root of tea times the square root of two TT. Okay, so this becomes the integral from 0 to 1 of two tea times one plus 40 square root of one plus 40 square d t plus, um, and then we have, um, the square root of two times integral from 0 to 1 of t plus t to the 1/2 tt. Okay, so the first thing a girl we're gonna have to do a, um you said so you is equal to one plus 40 squared. D'you is equal to eight t t t um, And so 1/4 d'you is equal to two t t t. So the first inter goal becomes 1/4 time's the integral of you to the 1/2 to you, and we're gonna go ahead and change our lower and upper bounds. And so when t zero us one and when he is one us five, um, and then this is gonna be a course, Um, plus the square root of two times integral from 0 to 1 of tea, please. Tea to the 1/2 TT. Okay. And so let's go ahead and integrate those. So this is gonna be 1/4 times 2/3 you to the three halves. And it went from 5125 Plus, then we have, um, square root of two over to t squared from and evaluated at one and zero, and then we're gonna have a plus two times escorted to over three t to the three. Has home once again evaluated at one in zero. So this becomes a 16 times, um, five to the 3/2 minus one. Um, plus the square root of two over, too. Plus two times of scrimmage to over three. And so, once again, if we put this over a common denominator, we get +16 times five to the three have minus one plus a seven times the square root of two

University of Central Arkansas