First, you must find the vector that is parallel to the line.

This vector is 

$\displaystyle v=\left \langle x_1-x_2,y_1-y_2,z_1-z_2\right \rangle$.

From the points we were given, this becomes 

$\displaystyle \left \langle -2,-4,-5\right \rangle$.

To form the parametric equations, we need to pick a point that lies on the line we want.

The point $\displaystyle \left ( 0,1,3\right )$ is used.

The vector form of the line is from the following equation 

$\displaystyle r=\left ( x_1,y_1,z_1\right )+t\left \langle v\right \rangle=\left ( 0,1,3\right )+t\left \langle -2,-4,-5\right \rangle$.

We then rewrite each expression in terms of the variables x, y, and z. 

Evaluate the line integral $\displaystyle \int_C f(x,y)ds$using the function 

$\displaystyle f(x,y)=y+\cos(\pi x)$ over the line segment $\displaystyle C$ from $\displaystyle (0,0)$ to $\displaystyle (4,5)$

Define the Parametric Equations to Represent $\displaystyle C$

The points given lie on the line $\displaystyle y = \frac{5}{4}x$. Define the parameter $\displaystyle x = t$, then $\displaystyle y$ can be written $\displaystyle y =\frac{5}{4}t$. Therefore, the parametric equations for $\displaystyle C$ are: 

$\displaystyle x(t)=t$

$\displaystyle y(t)=\frac{5}{4}t$

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 The line integral of a function $\displaystyle f(x,y)$ along the curve $\displaystyle C$ with the parametric equation $\displaystyle x(t)$ and $\displaystyle y(t)$ with $\displaystyle a\leq t\leq b$ is defined by: 

$\displaystyle \int_C f(x,y)ds = \int_a^bf(x(t),y(t))\left \| \vec{r'}(t)\right \|dt$                                  (1)

Where $\displaystyle \vec{r'(t)}$ is the vector derivative of the vector $\displaystyle \vec{r}(t)=\left< x(t),y(t)\right>$, therefore $\displaystyle \left \| \vec{r'(t)}\right \|$ is simply the magnitude of the vector derivative. 

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Write the vector $\displaystyle \vec{r(t)}$:

$\displaystyle \vec{r(t)}=\left< t, \frac{5}{4}t\right>$

Differentiate, 

$\displaystyle \vec{r'(t)}=\left< 1,\frac{5}{4}\right>$

The absolute value (magnitude) of this vector is: 

$\displaystyle \left \| \vec{r'}(t)\right \|=\left \| \left< 1,\frac{5}{4 }\right>\right \|=\sqrt{1^2 +\left(\frac{5}{4} \right )^2}=\frac{\sqrt{41}}{4}$

Write the function $\displaystyle f(x,y)$ in terms of the parameter $\displaystyle t$: 

$\displaystyle f(x(t),y(t))=f\left(t,\frac{5}{4}t\right) =\frac{5}{4}t+\cos(\pi t)$

Insert everything into Equation (1) noting that the limits of integration will be $\displaystyle 0\leq t \leq 4$ due to the fact that the parameter $\displaystyle x(t)=t$ varies from $\displaystyle 0$ to $\displaystyle 4$ over the line segment we are integrating over. 

$\displaystyle \int_C(y+\cos(\pi x))ds=\int_0^4\left(\frac{5}{4}t+\cos(\pi t)\right)\frac{\sqrt{41}}{4}dt$

$\displaystyle =\frac{\sqrt{41}}{4}\left[\frac{5}{8}t^2+\frac{1}{\pi}\sin(\pi t) \right ]_{t=0}^{t=4}$

$\displaystyle =\frac{\sqrt{41}}{4}\left[\frac{5}{8}(16)+\frac{1}{\pi }\sin(4 \pi) -0-\frac{1}{\pi}\sin(0)\right ]$

$\displaystyle =\frac{5\sqrt{41}}{2}$

Since there isn't a specific path we need to take, we just evaluate $\displaystyle f(x,y,z)$ at the end points.

$\displaystyle \int_{C} \nabla f \cdot d\vec{r}=f(1,1,-2)-f(3,4,5)$

$\displaystyle =\cos(2\pi (1))+\sin(3\pi (1))+(1)(1)(-2)-(\cos(2\pi (3))+\sin(4\pi (4))+(3)(4)(5))$

$\displaystyle =\cos(2\pi)+\sin(3\pi)-2-(\cos(6\pi)+\sin(12\pi)+60)$

$\displaystyle =1+0-2-(1+0+60)=-62$

First we need to make sure that the conditions for Green's Theorem are met. 

The conditions are met because it is positively oriented, piecewise smooth, simple, and closed under the region (see below).  

$\displaystyle 0\leq x \leq 1$

$\displaystyle 0 \leq y \leq 2x$

In this particular case $\displaystyle P=5xy$, and $\displaystyle Q=x^3y^2$, where $\displaystyle P$, and $\displaystyle Q$ refer to $\displaystyle \oint_C P \ dx +Q \ dy$.

We know from Green's Theorem that

$\displaystyle \oint_C P \ dx +Q \ dy=\int \int_D \Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big) dA$

So lets find the partial derivatives.

$\displaystyle \frac{\partial Q}{\partial x}=3x^2y^2$

$\displaystyle \frac{\partial P}{\partial y}=5x$

$\displaystyle \oint_C 5xy \ dx +x^3y^2 \ dy=\int \int_D 3x^2y^2-5x \ dA$

$\displaystyle =\int_{0}^{1} \int_{0}^{2x} 3x^2y^2-5x \ dy\ dx$

$\displaystyle =\int_{0}^{1} \Big(\frac{1}{3}\cdot3x^2y^3-5xy\Big)\Big|_{0}^{2x}\ dx$

$\displaystyle =\int_{0}^{1} \Big(x^2(2x)^3-5x(2x)\Big)\ dx$

$\displaystyle =\int_{0}^{1} 8x^5-10x^2\ dx$

$\displaystyle = \frac{4}{3}x^6-\frac{10}{3}x^3\Big|_{0}^{1}$

$\displaystyle = \frac{4}{3}(1)^6-\frac{10}{3}(1)^3-0=\frac{4}{3}-\frac{10}{3}=-\frac{6}{3}=-2$

Using Green's theorem

$\displaystyle \int_C{F}dr=\int \int\left[{\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}}\right]dA$

$\displaystyle \frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=-4-3=-7$

$\displaystyle Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

since the region is oriented clockwise, we would have

$\displaystyle \int_C{F}dr=-\int\int\left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

which gives us

$\displaystyle \int_C{F} dr=-(-7)\times4=28$

$\displaystyle \int_C 5y dx - 4x dy$

Using Green's theorem

$\displaystyle \int_C F dr=\int\int \left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

$\displaystyle Q=-4x\\ \frac{\partial{Q}}{\partial{x}}=-4$ $\displaystyle P=5y\\ \frac{\partial{P}}{\partial{y}}=5$

$\displaystyle Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

Since the region is oriented clockwise

$\displaystyle \int_C F dr=-\int\int (-4-5)dA$

$\displaystyle \int_C F dr=\int\int 9dA=(9)(4)=36$

In order to find the divergence, we need to remember the formula.

Divergence Formula:

$\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$, where $\displaystyle P$, $\displaystyle Q$, and $\displaystyle R$ correspond to the components of a given vector field $\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$.

Now lets apply this to our situation.

$\displaystyle div\vec{F}=\frac{\partial}{\partial x}\Big(yz\ln(x)\Big)+\frac{\partial}{\partial y}\Big(2xyz\Big)+\frac{\partial}{\partial z}\Big(z\Big)$

$\displaystyle div\vec{F}=\frac{yz}{x}+2xz+1$

In order to find the divergence, we need to remember the formula.

Divergence Formula:

$\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$, where $\displaystyle P$, $\displaystyle Q$, and $\displaystyle R$ correspond to the components of a given vector field $\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$.

Now lets apply this to our situation.

$\displaystyle div\vec{F}=\frac{\partial}{\partial x}\Big(x^2z\ln(y)\Big)+\frac{\partial}{\partial y}\Big(x2^{y^2}z\Big)+\frac{\partial}{\partial z}\Big(x^x\Big)$

$\displaystyle div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}+0$

$\displaystyle div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}$

In order to find the divergence, we need to remember the formula.

Divergence Formula:

$\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$, where $\displaystyle P$, $\displaystyle Q$, and $\displaystyle R$ correspond to the components of a given vector field $\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$.

Now lets apply this to our situation.

$\displaystyle div\vec{F}=\frac{\partial}{\partial x}\Big(xy\Big)+\frac{\partial}{\partial y}\Big(xz\Big)+\frac{\partial}{\partial z}\Big(xyz\Big)$

$\displaystyle div \vec{F}=y+0+xy$

$\displaystyle div \vec{F}=y+xy=y(1+x)$

In order to find the divergence, we need to remember the formula.

Divergence Formula:

$\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$, where $\displaystyle P$, $\displaystyle Q$, and $\displaystyle R$ correspond to the components of a given vector field $\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$.

Now lets apply this to our situation.

$\displaystyle div\vec{F}=\frac{\partial}{\partial x}\Big(xyz\Big)+\frac{\partial}{\partial y}\Big(x^2z^{-4}\Big)+\frac{\partial}{\partial z}\Big(y^{-10}z^3\Big)$

$\displaystyle div\vec{F}={yz}+0+3y^{-10}z^2=yz+\frac{3z^2}{y^{10}}$