Find the value of integral $\int_{C}\left(z^{2}+x y\right) d s$, where $C$ is parmeterized by $\vec{r}(t)=\langle-3 t, 5 t,-t\rangle$ for $0 \leq t \leq 6$ Question Help: G VIDEO D Post to forum

Step 1 ：First, we need to understand the problem. We are asked to find the value of the integral $\int_{C}\left(z^{2}+x y\right) d s$, where $C$ is parameterized by $\vec{r}(t)=\langle-3 t, 5 t,-t\rangle$ for $0 \leq t \leq 6$.

Step 2 ：Next, we need to express $z^{2}+x y$ in terms of $t$. We know that $x=-3t$, $y=5t$, and $z=-t$. So, $z^{2}+x y=(-t)^{2}+(-3t)(5t)=t^{2}+15t^{2}=16t^{2}$.

Step 3 ：Now, we need to find $ds$. We know that $ds=|d\vec{r}|=|\vec{r}'(t)|dt$, where $\vec{r}'(t)$ is the derivative of $\vec{r}(t)$. We have $\vec{r}'(t)=\langle-3, 5,-1\rangle$, so $|\vec{r}'(t)|=\sqrt{(-3)^{2}+(5)^{2}+(-1)^{2}}=\sqrt{35}$.

Step 4 ：Therefore, $ds=\sqrt{35}dt$.

Step 5 ：Substitute $16t^{2}$ for $z^{2}+x y$ and $\sqrt{35}dt$ for $ds$ in the integral, we get $\int_{C}\left(z^{2}+x y\right) d s=\int_{0}^{6}16t^{2}\sqrt{35}dt$.

Step 6 ：Now, we can calculate the integral. We have $\int_{0}^{6}16t^{2}\sqrt{35}dt=16\sqrt{35}\int_{0}^{6}t^{2}dt=16\sqrt{35}\left[\frac{1}{3}t^{3}\right]_{0}^{6}=16\sqrt{35}\left(\frac{1}{3}(6)^{3}-0\right)=16\sqrt{35}\times72=1152\sqrt{35}$.

Step 7 ：Finally, we check our result. The integral $\int_{C}\left(z^{2}+x y\right) d s$ is indeed equal to $1152\sqrt{35}$, which is in its simplest form.

Step 8 ：So, the value of the integral $\int_{C}\left(z^{2}+x y\right) d s$ is $\boxed{1152\sqrt{35}}$.