Like

Report

In Exercises $43 - 46 ,$ use a CAS to perform the following steps to evaluate the line integrals.

$$

\begin{array} { l } { \text { a. Find } d s = | \mathbf { v } ( t ) | d t \text { for the path } \mathbf { r } ( t ) = g ( t ) \mathbf { i } + h ( t ) \mathbf { j } + k ( t ) \mathbf { k } \text { . } } \\ { \text { b. Express the integrand } f ( g ( t ) , h ( t ) , k ( t ) ) | \mathbf { v } ( t ) | \text { as a function of the parameter } t . } \\ { \text { c. Evaluate } \int _ { C } f d s \text { using Equation } ( 2 ) \text { in the text. } } \end{array}

$$

$$

\begin{array} { l } { f ( x , y , z ) = x \sqrt { y } - 3 z ^ { 2 } ; \quad \mathbf { r } ( t ) = ( \cos 2 t ) \mathbf { i } + ( \sin 2 t ) \mathbf { j } + 5 t \mathbf { k } } \\ { 0 \leq t \leq 2 \pi } \end{array}

$$

A. $\sqrt{29} d t$

B. $\int_{0}^{2 \pi}\left(\cos 2 t \sqrt{\sin 2 t}-75 t^{2}\right) \sqrt{29} d t$

C. $-200 \sqrt{29} \pi^{3}$

You must be signed in to discuss.

Missouri State University

Oregon State University

Harvey Mudd College

Baylor University

Okay. What we want to go ahead and do is we're given a function, uh, X, Why Z which is equal to X time to square with a Y minus three z squared. And, um, the parameter ized is R of T is equal to co sign of to t I plus sign of to t j plus five t k where t is going to go from 0 to 2 pi inclusive, okay? And we want eventually set up the integral. So we want to go ahead and set up the integral role over the curve of f of eggs. Why Z d s okay. And then we want to do this in our calculator. So the first thing we need to do is find d s. And so we know D s is dependent upon this V of tea, which is gonna be the derivative of r of t. So this is gonna be negative to sign of tea. I plus to co sign of T J plus five K. And so the magnitude of e of t is the square root of a native to sign of that should have a two in there to tea and where a square that plus to co sign of two t square That plus 25 Because we scared squared the 25 the five And so this will end up as a square with 29. So d s is the square root of 29 TT. And so now the internal integral becomes 0 to 2 pi, and now we're transfer or function and ex wives and seize tor function in teas. And so we know that co sign, um, X is co sign of to t. Why is, um, sign of to tea and we put it under the square root. Um and then this is minus three times five T square, which gives me a minus 75 t squared, and that is gonna be multiplied by the square to 29. And we're integrating with respect to tea. And so we're gonna go ahead and we want to put this in our calculator. Um, I actually have a online a t. I 89 online simulator. I've actually already put it into practice. I had some issues with parentheses, of course. And so what I've done is I have already put this in, so I have the integral. And of course we had to put some. Prentiss is in there and, of course, were integrating with respect to X on her calculator. And so we have co side of two eggs times this crow decided to eggs minus the 75 X squared and times that square to 29. And, of course, we're integrating with respect eggs from 0 to 2 pi. And so when I do that, we actually get a negative 200 times pi cubed. Time to square 2 29 Now, for some reason, on my 84 I was getting, um, non riel answers. And so you probably need to go to, uh, her higher level system there are on Wolfram. There is, um, integration simulator. There's multiple integration simulators out there. Or you can have that tea I 89 if you're having difficulty on just a 84 or I don't have an inspire. So you might want to check out the inspire a swell, um And so this was, um, equal. Thio Negative 200 hi Q Times Square to 29

University of Central Arkansas