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Problem 15 Medium Difficulty

The paths of integration for Exercises 15 and 16
Integrate $f ( x , y , z ) = x + \sqrt { y } - z ^ { 2 }$ over the path from $( 0,0,0 )$ to $( 1,1,1 )$ (see accompanying figure) given by
$$
\begin{array} { l l } { C _ { 1 } : } & { \mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } , \quad 0 \leq t \leq 1 } \\ { C _ { 2 } : } & { \mathbf { r } ( t ) = \mathbf { i } + \mathbf { j } + t \mathbf { k } , \quad 0 \leq t \leq 1 } \end{array}
$$

Answer

$\frac{1}{6}(5 \sqrt{5}+9)$

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Video Transcript

Okay. What we want to do is step through the process of being able to integrate. Um, the function f of X y Z, which is equal to X plus the squared Y minus C squared over the path from 00021 11 Um, and it's not gonna be just a straight line path is going to be two separate curved pass. And I'm gonna go ahead and show all the picture and so is going to start at 00 And the first curve, um, goes from 00 to a 0.1 comma, one color zero, and then the second curve is thin. Um, going to go up to the 0.11 one. Okay. And so the first curve is defined to be our t. The path is to find B r t is equal to t I plus T Square, J and T is going to go from 0 to 1 inclusive. And, um, the second path is our t is equal toe I plus J plus K t k. Sorry. Um, and once again, it's Syria. Tha one inclusive is t. Okay. So what we do know is that, um, the integral, um, over the curve of the function. Um, with respect to d s and so is equal to what we can do is actually ads are two pieces together, so it could be the integral over the first curve of, or function related to that first curve plus the integral over the second curve related to that function. Um, based off of that second, Curtis. Well, okay. And so you noticed we have this d s in here? Well, in all of them and d s, um, is actually, um to find to be the magnitude of this v of t d t. Okay, so we gotta find this magnitude of e of T. Well, V A t is actually the derivative of r of T. And so we're gonna go back up here and take the derivative of both ours of teas. And so this becomes an eye plus a to T. J. And so the magnitude is equal to the square root of whatever is in front of I. So this could be a one square plus whatever is in front of the J terms of two t squared, plus whatever is in front of the K term, which we don't have. So this is gonna be the square root of one plus four t squared. And so D s for the first curb is the square root of one plus four T square. DT. Okay, Now, for the second curve, V of T is equal to, um are teased derivatives. So this is gonna be equal to, um, zero plus zero, plus que and so the magnitude is gonna be the square root of one squared, which is one. So for the second curve, D s is exactly equal to d. T. And so let's go ahead. You can continue. I'm going to go ahead and write our function once again. So this was X plus the square toe. Why minus z squared and I'm gonna go ahead and write both or values. Um, our keys and that waas, um, for the first curve was t I placed T Square J. And for the second curve, it waas um, I plus J plus t k. Okay, So the integral of the first integral over the first curve goes from 0 to 1 of our function is defined by that first curve. And so this is gonna be t um because our X values t based off of our r of t um, plus the square root of t squared plus zero. And then this is gonna be times that square root of one plus four T square. DT Um, plus now we're doing our second curve in fear to one. And this time our function is going to be one or ex Values can be won plus a why value of one minus z squared. Um, t squared. I'm sorry, d t. Because that was our d s. Okay, so this is gonna be the integral from 0 to 1 of, um to t times the square root of one plus 40 squared. Um, plus, seeing a girl from 0 to 1 of two minus t squared t t I'm gonna go ahead. And And this was t squared. Ok, now on the first in a world, we're gonna go ahead and do a use up, and so we're gonna let you be one plus four t squared. So d'you is equal to a t d t. I only have a to t d t inside that integral. So I'm gonna divide out that one. That four. So I have 1/4 to use equal to two to DT. So this first inter goal becomes, and then we're gonna go ahead and change our upper or lower limits. So when tea is zero, you is one when tea is one US five and we have the 1/4 you to the 1/2 do you? And then we have that, plus the integral from 0 to 1 of two money's t squared T t. Okay, so this becomes 1/4 times 2/3 you to the three halfs, and we're gonna evaluate it at five. And at one plus two tea, and we're gonna evaluate that at one and zero minus 1 30 cube evaluated at one in zero. Okay, so when we do that, we get, um this is gonna be equal toe 16 times five, Route five, um, minus one. And they wouldn't have the plus two minus 1/3. Okay. And then we're gonna go ahead and put everything of our common denominator and simplify. We should get 1/6 times five. Route five plus a nine

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