Green's theorem question needs checking

**Dr Zoidburg** · May 18th 2010, 05:14 PM

Question:
Use Green's theorem to evaluate the following line integrals. Assume that the curve is transversed counterclockwise.
(a) $\oint_{\gamma}3xy.\partial x + 2xy.\partial y,$ where $\gamma$ is the rectangle with corners at (-2,1),(-2,2),(4,1) & (4,2).

(b) $\oint_{\gamma}(x^2 - y).\partial x + x.\partial y,$ where $\gamma$ is the circle $x^2+y^2=4$

My answers:
(a) P=3xy & Q=2xy
Green's Theorem gives:
$\int \int_{D}(2y-3x).\partial A$
where D is the rectangle enclosed by $-2 \leq x \leq 4 \; and \; 1 \leq y \leq 2$
I found if I just go ahead and integrate I end up with 0 as the answer.
However, since the path is not smooth, should I take the integral along each side separately and then add them?
Doing this I end up with:
first side (-2,1) to (-2,2). Make x=-2 -> dx = 0. Make y=t -> dy=dt.
$\int_{(-2,1)}^{(-2,2)}3xy.\partial x + 2xy.\partial y = \int_{t=1}^{2}3(-2)t.0 + 2(-2)t.\partial t = \int_{1}^{2}-4t.\partial t = \bigg[ -2t^2 \bigg]_{1}^{2} = -6$
I did the same for the other 3 sides and got 12, 18 and 36. This makes the entire integral 60.
Does this look ok? Or do I need to take the absolute values of each, making the final answer 72.

(b) $P=x^2-y \; and \; Q=x$
Green's theorem gives:
$\oint_{\gamma } (x^2-y).\partial x + x.\partial y = \int \int_{D}2.\partial A$
where D is a disk of radius 4 centred at the origin.
Since this is a disk, is the best way to do this integral is using polar co-ordinates?
This is what I got:
$\int_{0}^{2\pi} \int_{0}^{4}2r.\partial r \partial \theta = \int_{0}^{2 \pi}\bigg[r^2\bigg]_{0}^{4}=\int_{0}^{2\pi}16.\partial\theta =\bigg[16\theta \bigg]_{0}^{2\pi}=32\pi$

**11rdc11** · May 18th 2010, 08:44 PM

Originally Posted by Dr Zoidburg

Question:
Use Green's theorem to evaluate the following line integrals. Assume that the curve is transversed counterclockwise.
(a) $\oint_{\gamma}3xy.\partial x + 2xy.\partial y,$ where $\gamma$ is the rectangle with corners at (-2,1),(-2,2),(4,1) & (4,2).

(b) $\oint_{\gamma}(x^2 - y).\partial x + x.\partial y,$ where $\gamma$ is the circle $x^2+y^2=4$

My answers:
(a) P=3xy & Q=2xy
Green's Theorem gives:
$\int \int_{D}(2y-3x).\partial A$
where D is the rectangle enclosed by $-2 \leq x \leq 4 \; and \; 1 \leq y \leq 2$
I found if I just go ahead and integrate I end up with 0 as the answer.
However, since the path is not smooth, should I take the integral along each side separately and then add them?
Doing this I end up with:
first side (-2,1) to (-2,2). Make x=-2 -> dx = 0. Make y=t -> dy=dt.
$\int_{(-2,1)}^{(-2,2)}3xy.\partial x + 2xy.\partial y = \int_{t=1}^{2}3(-2)t.0 + 2(-2)t.\partial t = \int_{1}^{2}-4t.\partial t = \bigg[ -2t^2 \bigg]_{1}^{2} = -6$
I did the same for the other 3 sides and got 12, 18 and 36. This makes the entire integral 60.
Does this look ok? Or do I need to take the absolute values of each, making the final answer 72.

(b) $P=x^2-y \; and \; Q=x$
Green's theorem gives:
$\oint_{\gamma } (x^2-y).\partial x + x.\partial y = \int \int_{D}2.\partial A$
where D is a disk of radius 4 centred at the origin.
Since this is a disk, is the best way to do this integral is using polar co-ordinates?
This is what I got:
$\int_{0}^{2\pi} \int_{0}^{4}2r.\partial r \partial \theta = \int_{0}^{2 \pi}\bigg[r^2\bigg]_{0}^{4}=\int_{0}^{2\pi}16.\partial\theta =\bigg[16\theta \bigg]_{0}^{2\pi}=32\pi$

a) Yes the answer is 0

b) change your limits for the radius to 0 and 2

**HallsofIvy** · May 19th 2010, 04:18 AM

Originally Posted by Dr Zoidburg

Question:
Use Green's theorem to evaluate the following line integrals. Assume that the curve is transversed counterclockwise.
(a) $\oint_{\gamma}3xy.\partial x + 2xy.\partial y,$ where $\gamma$ is the rectangle with corners at (-2,1),(-2,2),(4,1) & (4,2).

(b) $\oint_{\gamma}(x^2 - y).\partial x + x.\partial y,$ where $\gamma$ is the circle $x^2+y^2=4$

My answers:
(a) P=3xy & Q=2xy
Green's Theorem gives:
$\int \int_{D}(2y-3x).\partial A$
where D is the rectangle enclosed by $-2 \leq x \leq 4 \; and \; 1 \leq y \leq 2$
I found if I just go ahead and integrate I end up with 0 as the answer.
However, since the path is not smooth, should I take the integral along each side separately and then add them?
Doing this I end up with:
first side (-2,1) to (-2,2). Make x=-2 -> dx = 0. Make y=t -> dy=dt.
$\int_{(-2,1)}^{(-2,2)}3xy.\partial x + 2xy.\partial y = \int_{t=1}^{2}3(-2)t.0 + 2(-2)t.\partial t = \int_{1}^{2}-4t.\partial t = \bigg[ -2t^2 \bigg]_{1}^{2} = -6$
I did the same for the other 3 sides and got 12, 18 and 36. This makes the entire integral 60.
Does this look ok? Or do I need to take the absolute values of each, making the final answer 72.

Why did you do that? The problem did not ask you to actually integrate along the path nor did it ask you to do the integral both ways to check Green's theorem- it only asks you to use Green's theorem to find the integral around the path- to integrate over the area. That integral is, indeed, 0 which means the integral around the path must also be 0.

The fact that you ask about whether you take the absolute value makes me wonder how well you know the integral itself. On of the first things you should have learned about the integral is that $\int_a^b f(x)dx= \int_a^c f(x)dx+ \int_c^b f(x) dx$ . No absolute value magically show up in that!

Now, you can check Green's theorem by integrating around the path itself but, of course, Green's theorem says the two integrals are the same. You should get 0 again, not 60 or 72.
On the first side, from (-2, 1) to (-2, 2), yes, you can take x= -2, y= t so that dx= 0, dy= dt. Then you have $\int{t= 1}^2 (3)(-2)(t)(0)+ 2(-2)(t)dt= -4\int{1}^2 t dt= \left[-2t^2\right]_1^2= -2(4- 1)= -6$ , exacly what you got.

On the side from (-2, 2) to (4, 2), you can take x= t, y= 2 so that dx= dt, dy= 0. Then you have $\int_{t= -2}^4 (3)(t)(2)dt+ 2(t)(2)0= 6\int_{-2}^4 t dt$ $= 3\left[t^2\right]_{-2}^4= 3(16- 4)= 36$ , again, exactly what you got.

On the side from (4, 2) to (4, 1), you can take x= 4, y= t so that dx= 0, dy= dt. Then you have $\int_{t= 2}^{1} (3)(4)(t)0+ (2)(4)(t)dt= 8\int_2^{1} t dt$ $= 4\left[t^2\right]_2^{1}= 4(1- 4)= -12$ . That is the negative of what you got.

Because we were going from (4, 2) down to (4, 1) the integral is from 2 to 1 so the integral is f(1)- f(2). We could have avoided integrating from a larger number to a smaller by letting y= -t so that y= 2, t= -2, and when y= 1, t= -1. Then we would have $\int_{-2}^{-1}$ but we would also have dy= -dt so the entire integral would be multiplied by -1.

On the final leg, from (4, 1) to (-2, 1), the same thing happens. Let x= t, y= 1 so that dx= dt, dy= 0. Now we have $\int_{t= 4}^{-2} 3(x)(1)dt+ 2(x)(1)(0)=$ $\int_4^{-2} 3x dx= 3\left[x^2\right]_4^{-2}= (3/2)(4- 18)= -18$ .

The total is -6+ 36- 12- 18= 0.

(b) $P=x^2-y \; and \; Q=x$
Green's theorem gives:
$\oint_{\gamma } (x^2-y).\partial x + x.\partial y = \int \int_{D}2.\partial A$
where D is a disk of radius 4 centred at the origin.
Since this is a disk, is the best way to do this integral is using polar co-ordinates?
This is what I got:
$\int_{0}^{2\pi} \int_{0}^{4}2r.\partial r \partial \theta = \int_{0}^{2 \pi}\bigg[r^2\bigg]_{0}^{4}=\int_{0}^{2\pi}16.\partial\theta =\bigg[16\theta \bigg]_{0}^{2\pi}=32\pi$

Actually, since you are integrating a constant over a disk, the best way to do this is not to integrate at all but to multiply the constant by the area of the disk! And, as 11rdc11 points out, the circle given by $x^2+ y^2= 4$ has radius $\sqrt{4}= 2$ .

**11rdc11** · May 19th 2010, 07:42 AM

Originally Posted by Dr Zoidburg

Question:
Use Green's theorem to evaluate the following line integrals. Assume that the curve is transversed counterclockwise.
(a) $\oint_{\gamma}3xy.\partial x + 2xy.\partial y,$ where $\gamma$ is the rectangle with corners at (-2,1),(-2,2),(4,1) & (4,2).

(b) $\oint_{\gamma}(x^2 - y).\partial x + x.\partial y,$ where $\gamma$ is the circle $x^2+y^2=4$

My answers:
(a) P=3xy & Q=2xy
Green's Theorem gives:
$\int \int_{D}(2y-3x).\partial A$
where D is the rectangle enclosed by $-2 \leq x \leq 4 \; and \; 1 \leq y \leq 2$
I found if I just go ahead and integrate I end up with 0 as the answer.
However, since the path is not smooth, should I take the integral along each side separately and then add them?
Doing this I end up with:
first side (-2,1) to (-2,2). Make x=-2 -> dx = 0. Make y=t -> dy=dt.
$\int_{(-2,1)}^{(-2,2)}3xy.\partial x + 2xy.\partial y = \int_{t=1}^{2}3(-2)t.0 + 2(-2)t.\partial t = \int_{1}^{2}-4t.\partial t = \bigg[ -2t^2 \bigg]_{1}^{2} = -6$
I did the same for the other 3 sides and got 12, 18 and 36. This makes the entire integral 60.
Does this look ok? Or do I need to take the absolute values of each, making the final answer 72.

(b) $P=x^2-y \; and \; Q=x$
Green's theorem gives:
$\oint_{\gamma } (x^2-y).\partial x + x.\partial y = \int \int_{D}2.\partial A$
where D is a disk of radius 4 centred at the origin.
Since this is a disk, is the best way to do this integral is using polar co-ordinates?
This is what I got:
$\int_{0}^{2\pi} \int_{0}^{4}2r.\partial r \partial \theta = \int_{0}^{2 \pi}\bigg[r^2\bigg]_{0}^{4}=\int_{0}^{2\pi}16.\partial\theta =\bigg[16\theta \bigg]_{0}^{2\pi}=32\pi$

Here is another quick tip. Since this is a polar rectangle you can seperate your intergral no variable dependent on the other.

$\int_{0}^{2\pi} \int_{0}^{2}2r.\partial r \partial \theta = \bigg(\int^{2\pi}_{0} \partial \theta \bigg)\bigg(\int^{2}_{0} 2r \partial r \bigg)$

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