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Numerade Educator



Problem 18 Medium Difficulty

Scalar line integrals in the plane
a. Find a parametric description for $C$ in the form $\mathbf{r}(t)=\langle x(t), y(t)\rangle,$ if it is not given.
b. Evaluate $\left|\mathbf{r}^{\prime}(t)\right|$
c. Convert the line integral to an ordinary integral with respect to the parameter and evaluate it.
$\int_{C}(x y)^{1 / 3} d s ; C$ is the curve $y=x^{2},$ for $0 \leq x \leq 1$


A. $$\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle, 0 \leq t \leq 1$$
B. $$\sqrt{1+4 t^{2}}$$
C. $$\frac{1}{12}\left(5^{\frac{3}{2}}-1\right)$$


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Video Transcript

and red 18. Find the parish warm. Versi party's gonna be evaluated. Our promise t the main tier two, then part C. We're going to convert Ah, line it, girl into the ordinary and grow. And if I read it and c is gonna be the curve y equals X squared from data one. So starting in part, a objective is to find this form No curves. Ah, problem. Michael's X squared. So let's just let exit TVT and why have tea t squared then our t is gonna be t comity squared. Okay, that's party, part B. I want the magnitude of or prime. So taken is gonna be the derivative t, which is one squared plus derivative T square, which is two t that tickles Kurt of one plus four t squared. Could part C on a valued the integral Using this term you're the integral off may be of f rt tons magnitude barf t details Pretty even Tareen and be so plugging all the values that we have If our of tea does it go dio exit e he his wife t by one third That's going to go toe tee Times t squared by one third, which is a good tea to know that we have this part, which is X Y two, part of third. Then we put that back into the into grow and grow from 0 to 1 of tea. So that's the first part multiplied by this parts the magnitude of our prime tea. All right. And this this integral we're gonna do you substitution because you can see the inside the derivatives gonna fit the outside. So taking you as one plus for T Square plus one then do you is good a t d t. And this matches this part right here. Now we do this, and we gotta change the limits. T equals zero. Then is you goes Warren. Plugging in zero into that t equals one than you equals five. Our new in the role in a row from 1 to 5. Uh, square to you Put the 1/8 do you? No, I felt reading the integral. We got one a times squared of you. There are three over half plus one. Evaluate that from 1 to 5. Uh, one a times 2/3 times. You three or two evaluated at 15 Right now we plug and our values for one in five and remotes by this part. Out we get 1/12 times. Five to three. Over two minus 12 3/2 in that equals one or 12. It sounds 53 or two minus one, which is equal to our line, integral.

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