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

a. Find $d s=|\mathbf{v}(t)| d t$ for the path $\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+$ $k(t) \mathbf{k} .$

b. Express the integrand $f(g(t), h(t), k(t))|\mathbf{v}(t)|$ as a function of the parameter $t .$

c. Evaluate $\int_{C} f d s$ using Equation $(2)$ in the text.

$$f(x, y, z)=\left(1+\frac{9}{4} z^{1 / 3}\right)^{1 / 4} ; \quad \mathbf{r}(t)=(\cos 2 t) \mathbf{i}+(\sin 2 t) \mathbf{j}+ t^{5 / 2} \mathbf{k}, \quad 0 \leq t \leq 2 \pi$$

## Discussion

## Video Transcript

Okay, so we're starting with a slender metal arch that is going to be denser at the bottom, then the top. And it lies along, um, why squared plus c squared equal toe one where Z is greater than zero. Um, and we also know that that density function is given by, um, to minus Z. And what we want to dio is find the moment of inertia about the X axis, and so we know that is equal to the integral over the curve of the function. Um, times the density function, um, with respect to s. Okay, so this is what we have. Um, and I think it's not every time to function. I apologize for that. Um, the moment of inertia is actually about the ex is gonna be Why squared plus c squared times the density, Um, and integrating with respect to ass. So the first thing we need to do is to find all these terms while we know this is we know this is a one. We know that the density function is to minus c and the D s is equal to the magnitude of e of t times t t and we also need to put find on the density function in terms of tea as well. So we know that, um, r of t is gonna be given, um, this it lies along is a circle in the y Z plane with a radius of one. So that's gonna be given by co sign of tea in Parametric form Times J plus sign of tea. Times K. Um, and we are so Z is bigger than zero. Then we know it's going to go from, um zero t will go from zero to pie because it's going to be above that y axis. Okay, um and so we know that, um, the derivative of our it gives me that V f t function. So that's gonna be native sign of tea times, J um plus co sign of tea times K. And so the magnitude of the of tea is gonna be the square root of negative sign of t squared plus co sign of tea squared, which is basically going to give me a one so d s is gonna be equal to DT. Okay, um and so let's go ahead and keep going. So that moment of inertia is equal toothy, integral from zero to pie. Uh, why square Cassie's square, Which is a one, um, times the density and the density Waas on two minus z. Well, Z is, um, sign of tea. So this is gonna be to minus sign of tea. Um, with respect to t. So we're gonna integrate this because a to t evaluated at pie and it zero and plus a coast on a T evaluated at pie and zero as well. So this gives me to pie minus zero groups that should be a minus. Plus, um, a negative one for co sign of pie minus coast on a zero, which is a one. And so this gives me two pi minus two. So there is that moment of inertia about the and it should be about the eggs axis, not the Z axis.

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