# Evaluate the line integral by two methods: (a) directly and (b) using Green’s Theorem

Evaluate the line integral by two methods: (a) directly and (b) using Green’s Theorem So I thought I knew how to do this problem but when I did it directly, the areas I got for each line were 0+2/3+4, but the overal area in the answer key is 2/3. I double checked the entire process twice when I got the 4. Maybe I'm parameterizing it wrong or used the wrong boundaries, but I was sure it was supposed to be from 0 to 1 when you parameterize in terms of t.

my work:

r(t1) = {0,0} + t{1,0}

= {t,0}

x=t, y=0

dx=1dt, dy=0

so integral C = 0

r(t2) = {1,0} + t{0,2}

= {1,2t}

x=1, y=2t

dx=0, dy=2dt

so integral C2 = integral C(0+8t^3*2)dt from 0 to 1 = 4

r3(t) = {1,2} + t{-1,-2}

={1-t,2-2t}

x=1-t, y=2-2t

dx = -1dt, dy = -2dt

integral c3 = integral c3( (1-t)(2-2t)(-1)dt + (1-t)^2(2-2t)^3(-2)dt

the left integral of this evaluated to 2/3 and the right I used an algebra calculator to simplify and it evaluated to 0. So the whole thing came out to 2/3 + 4, but its really 2/3.

Also, when I used green's theorem I ended up getting 6, but again its still supposed to be 2/3. I used the double integral of (dQ/dx -dP/dy)dydx with y going from 0 to 2x ( i got 2x with y=mx+b) and x going from 0 to 1. and I get 6. Am I doing something fundamentally wrong? Can someone show me every detail so I can see where I went wrong? Thanks.

my work:

double integral(dQ/dx -dP/dy)dydx =

integral(0 to 1)integral(0 to 2x)(2xy^3 - x)dxdy =

2xy^4/4 -xy ]2x to 0

2x(2x)^4 - 2x^2

8x^5 -2x^2 ] 1 to 0

= 6

• Can you please post your work...so we can see what you did. Thanks. – Ahmed S. Attaalla Dec 4 '17 at 3:19
• i put it up, itll take me a while to find the latex commands and apply them all – 2316354654 Dec 4 '17 at 4:02

## Using Green's Theorem:

The vector field we are dealing with is $F(x,y)=(xy,x^2y^3)$. The region you described is correct, $x$ goes from $0$ to $1$ and $0\leq y\leq 2x$.

By Green's Theorem we have:

$$I=\int_{0}^{1}\int_{0}^{2x}\left(\frac{d(x^2y^3)}{dx}-\frac{d(xy)}{dy}\right)dydx=\int_{0}^{1}\int_{0}^{2x}(2xy^3-y)dydx$$ You can evaluate this integral and the result is $\frac{2}{3}$.

## By definition

We take $A=(0,0), B= (1,0), C=(1,2)$. Then by using a line parametrization we have:

$$\begin{gather} r_1(t) = t(1,0)+(1-t)(0,0)=(t,0)\\ r_2(t) = t(1,2)+(1-t)(1,0)=(1,2t)\\ r_3(t) = t(0,0)+(1-t)(1,2)=(1-t,2(1-t)) \end{gather}$$

We know that the line integral $I$ is equal to:

$$I=\sum_{i=1}^3\int_0^1\langle F(r_i(t)),r_i'(t)\rangle dt$$

With

$$\begin{gather} I_1 = 0 \\ I_2 = \int_0^1\langle (2t,8t^3),(0,2)\rangle dt=16\int_0^1t^3=4\\ I_3 = \int_0^1\langle (2(1-t)^2,8(1-t)^5),(-1,-2)\rangle dt \end{gather}$$

And that last integral evaluates to $\frac{-10}{3}$. It follows $4-\frac{10}{3}=\frac{2}{3}$.

• shouldn't d(xy)/dy be x? – 2316354654 Dec 4 '17 at 4:27
• Yes! Sorry I didn't see that typo. It still evaluates to $\frac{2}{3}$ – Ignacio Rojas Dec 4 '17 at 4:30
• Can you explain how you set up the integral from the vector notation in I3? – 2316354654 Dec 4 '17 at 4:43
• When evaluating a line integral by definition, you take $I=\int_0^1\langle F(r(t)),r'(t)\rangle dt$ where $r$ is the parametrization for the line segment from $C$ to $A$. This means that $I_3=\int_0^1\langle F(r_3(t)),r_3'(t)\rangle dt$. – Ignacio Rojas Dec 4 '17 at 4:45
• can you show me how you went from that vector version to the actual integral that was evaluated? – 2316354654 Dec 4 '17 at 4:47