91影视

It is given that $\mathbf{F}(x,y,z)=\frac{\ln (z+4)}{x}\mathbf{i}+xyz\mathbf{j}+y\mathbf{k}$

The curve is parametrized by $x=4t,y=t^2,z=1-t$for$0鈮�t鈮�1$

The curve is parametrized as

for $0鈮�t鈮�1$

${\mathbf{r}}^{\text{'}}\left(t\right)=\frac{d}{dt}\mathbf{r}\left(t\right)$

$=鉄�4,2t,-1鉄�$

The vector field is written as

$\mathbf{F}(x,y,z)=\frac{\ln (z+4)}{x}\mathbf{i}+xyz\mathbf{j}+y\mathbf{k}$

$x=4t,y=t^2,z=1-t$

It becomes

$\mathbf{F}(x,y,z)=\frac{\ln (z+4)}{x}\mathbf{i}+xyz\mathbf{j}+y\mathbf{k}$

$\mathbf{F}\left(x\right(t),y(t),z(t\left)\right)=\frac{\ln \left(\right(1-t)+4)}{4t}\mathbf{i}+4t路t^2(1-t)\mathbf{j}+t^2\mathbf{k}$

$=\frac{\ln (5-t)}{4t}\mathbf{i}+4t^3(1-t)\mathbf{j}+t^2\mathbf{k}$

The required integral is

${鈭�}_C\mathbf{F}(x,y,z)路d\mathbf{r}$

$={鈭�}_C\mathbf{F}\left(x\right(t),y(t),z(t\left)\right)路{\mathbf{r}}^{\text{'}}\left(t\right)dt$

Using values

$={鈭�}_0^1\left[\left(\frac{\ln (5-t)}{4t}\right)\left(4\right)+\left(4t^3(1-t)\right)\left(2t\right)+\left(t^2\right)(-1)\right]dt$

Using dot product

$={鈭�}_0^1\left[\frac{\ln (5-t)}{t}+8t^4(1-t)-t^2\right]dt$

$={鈭�}_0^1\left[\frac{\ln (5-t)}{t}+8t^4-8t^5-t^2\right]dt$

$={鈭�}_0^1\left[\frac{\ln (5-t)}{t}\right]dt+{鈭�}_0^1\left(8t^4-8t^5-t^2\right)dt$

$={鈭�}_0^1\left[\frac{\ln (5-t)}{t}\right]dt+{\left[\frac{8t^5}{5}-\frac{8t^6}{6}-\frac{t^3}{3}\right]}_0^1$

$={鈭�}_0^1\left[\frac{\ln (5-t)}{t}\right]dt+\left(\frac{8(1{)}^5}{5}-\frac{8(1{)}^6}{6}-\frac{1^3}{3}\right)-\left(\frac{8(0{)}^5}{5}-\frac{8(0{)}^6}{6}-\frac{0^3}{3}\right)$

$={鈭�}_0^1\left[\frac{\ln (5-t)}{t}\right]dt+\left(-\frac{1}{15}\right)-0$

$=-\frac{1}{15}+{鈭�}_0^1\left[\frac{\ln (5-t)}{t}\right]dt$

Hence, the line integral is

${鈭�}_C\mathbf{F}(x,y,z)路d\mathbf{r}=-\frac{1}{15}+{鈭�}_0^1\left[\frac{\ln (5-t)}{t}\right]dt$