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

enter image description here

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}


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

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

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


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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.