# Math Help - Line integrals. Please help

1. ## Line integrals. Please help

Im not completely getting line integrals. Can you please help me with these questions.

I get 3 for (a) and (b) is that right?

Not sure how to do this one

I was thinking $\displaystyle\int_C F \cdot dr = \displaystyle\int_C F \cdot T ds = \displaystyle\int_C |F|cos \theta$
$|F| = \sqrt {x^4y^2 + 16}$. Not sure about theta. Is any of that the right thing to do?

I get $\displaystyle \frac{\partial f}{\partial y} = -e^xsiny$ and $\displaystyle \frac{\partial g}{\partial x} = -e^xcosy$. Not sure what to do next or if that is right.

For green's theorem it would be $\displaystyle\int \displaystyle\int_R (2x-2y) dA$? But im not sure about the limits. As for doing it directly, not sure about that either.

Please help. I really need to know this for my final exam

2. anyone please?

3. (1) Along the line from (0, 0) to (1, 1) y= x and dy= dx.
$\int_C (3x+ 2y)dx+ (2x- y)dy= \int_0^1 (3x+ 2x)dx+ (2x- x)dx$ $= \int_0^1 6x dx= \left[3x^2\right]_0^1= 3$.

Along $y= x^2$, dy= 2x dx.
$\int_C (3x+ 2y)dx+ (2x- y)dy= \int_0^1 (3x+ 2x^2)dx+ (2x- x^2)(2xdx)$ $= \int_0^1 3x+ 6x^2- 2x^3 dx= \left[\frac{3}{2}x^2+ 2x^3- \frac{1}{2}x^4\right]_0^1= 3$.

Yes, both (a) and (b) are "3".

Or we could have noted that $(3x+ 2y)_y= 2= (2x- y)_y$ so this integral is independent of the path and not have had to do the second integral.

(2) The curve, $x^2+ y^2= 1$ is, of course, a circle of radius 1 with center (0, 0). A standard parameterization for that is $x= cos(\theta)$, $y= sin(\theta)$ for which $dx= -sin(\theta)d\theta$ and $dy= cos(\theta)d\theta)$.

Of course, (1, 0)= (cos(0), sin(0)) and $(0, 1)= (cos(\pi/2), sin(\pi/2))$.

Now, we have $(x^2+ y^2)dx- x dy= (1)(-cos(\theta) d\theta)- cos(\theta)(sin(\theta)d\theta)$ and so $\int_C (x^2+ y^2)dx- x dy= \int_0^{\pi/2} (-cos(\theta)- cos(\theta)sin(\theta))d\theta$ $= -\int_0^{\pi/2}(1+ sin(\theta))cos(\theta) d\theta$ which can be integrated with the substitution $u= sin(\theta)$.

(3) Don't use " $F\cdot T= |F|cos(\theta)$", use " $(a, b)\cdot (x, y)= ax+ by$".

With $\vec{r}= e^t\vec{i}+ e^{-t}\vec{j}$, $d\vec{r}= e^t\vec{i}- e^{-t}\vec{j}$ and $\vec{F}= x^2y\vec{j}+ 4\vec{j}$ so $\vec{F}\cdot d\vec{r}= (e^{2t})(e^{-t})\vec{i}+ 4\vec{j}= e^t\vec{i}+ 4\vec{j}$.

So $\vec{F}\cdot d\vec{r}= (e^{2t}- 4e^{-t})dt$.

$\int_C \vec{F}\cdot d\vec{r}= \int_0^1 (e^{2t}- 4e^{-t})dt$.

(4) $\vec{F}(x,y)= e^x cos(y)\vec{i}- e^x sin(y)\vec{j}$. That will be a "conservative vector field" if and only if $\frac{\partial e^x cos(y)}{\partial y}= \frac{\partial -e^x sin(y)}{\partial x}$

Yes, $\frac{\partial e^x cos(y)}{\partial y}= -e^x sin(y)$ but $\frac{\partial -e^x sin(y)}{\partial y}$ is NOT equal to $-e^x cos(y)$ as you have. You differentiated with respect to y again! Since the derivative of $e^x$ is $e^x$, $\frac{\partial -e^x sin(y)}{\partial y}= -e^x sin(y)$.

Now, since those are the same, this is a "conservative vector field". That means that there is a function F(x,y) such that $\nabla F= \frac{\partial F}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}= e^x cos(y)\vec{i}- e^x sin(y)\vec{j}$. That is, $\frac{\partial F}{\partial x}= e^x cos(y)$ and $\frac{\partial F}{\partial y}= -e^x sin(y)$.

Since $\frac{\partial F}{\partial x}= e^x cos(y)$ and the integral of $Ae^x$ is $Ae^x+ C$, we must have $F(x,y)= e^x cos(y)+ g(y)$ where g can be any differentiable function of y- since the partial derivative with respect to x "treats y like a constant", the "constant of integration" can, in fact, be any function of y.

But knowing that $F(x,y)= e^x cos(y)+ g(y)$ we know that $\frac{\partial F}{\partial y}= -e^x sin(y)+ \frac{dg}{dy}$. And since we also have $\frac{\partial F}{\partial y}= -e^x sin(y)$, we must have $-e^x sin(y)+ \frac{dg}{dy}= -e^x sin(y)$ so $\frac{dg}{dy}= 0$.

That is, g really is a constant and $F(x,y)= e^x cos(y)+ C$.

(5) Yes, by Green's theorem $\oint_C y^2dx+ x^2 dy= \int\int_R (2x- 2y) dA$ where R is the region bounded by path C.

Since C is "the square with vertices (0, 0), (1, 0), (1, 1), and (0, 1)", R is the interior of that square. The limits of integration are x:0 to 1, y:0 to 1
$2\int_{x=0}^1\int_{y=0}^1 (x- y)dy dx$.

To do this "directly", since there are corners at (0, 0), (1, 0), (1, 1), and (0, 1) (the path is not "smooth"), take the integrals along the four sides separately and add.

For example, along the line from (0, 0) to (1, 0), we can take x= t, y= 0 so dx= dt, dy= 0. $\int_{(0,0)}^{(1,0)} y^2 dx+ x^2dy= \int_{t=0}^1 0^2 dx+ x^2 0= 0$.

Along the line from (1, 0) to (1, 1), we can take x= 1, y= t so dx= 0, dy= dt. $\int_{(1,0)}^{(1,1)} y^2dx+ x^2dy= \int_{t=0}^1 t^2(0)+ 1^2 dt= \int_{t=0}^1 dt= 1$.

You finish the last two lines: from (1, 1) (0, 1) by letting x= 1- t, y= 0, and from (0, 1) to (0, 0) by letting x= 0, y= 1- t.

4. thanks so much