Find the value of integral int_{C}left(x^{2}+y^{2}+zright) d s, where C is parmeterized by vec{r}(t)=langle 3 cos (4 t), 3 sin (4 t), 2 trangle for 0 leq t leq 4.

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Step:1 ：Let&#x27;s parameterize the given curve C by the vector function (vec{r}(t)=langle 3 cos (4 t), 3 sin (4 t), 2 trangle) for (0 leq t leq 4).Step:2 ：Substitute the parameterized values of x, y, and z into the function. So, x = 3*cos(4*t), y = 3*sin(4*t), and z = 2*t.Step:3 ：Substitute these values into the function to get (f = 2*t + 9*sin(4*t)^{2} + 9*cos(4*t)^{2}).Step:4 ：Integrate this function over the given limits of t from 0 to 4.Step:5 ：The value of the integral is 52.

Answer

Step: 1 ：Let&#x27;s parameterize the given curve C by the vector function (vec{r}(t)=langle 3 cos (4 t), 3 sin (4 t), 2 trangle) for (0 leq t leq 4).Step: 2 ：Substitute the parameterized values of x, y, and z into the function. So, x = 3*cos(4*t), y = 3*sin(4*t), and z = 2*t.Step: 3 ：Substitute these values into the function to get (f = 2*t + 9*sin(4*t)^{2} + 9*cos(4*t)^{2}).Step: 4 ：Integrate this function over the given limits of t from 0 to 4.Step: 5 ：The value of the integral is 52.

Find the value of integral $\int_{C}\left(x^{2}+y^{2}+z\right) d s$, where $C$ is parmeterized by $\vec{r}(t)=\langle 3 \cos (4 t), 3 \sin (4 t), 2 t\rangle$ for $0 \leq t \leq 4$.

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