Verify Green's Theorem in the given problem

[chwala]

Homework Statement

See attached

Relevant Equations

Green's theorem

My lines are as follows;

$y=\sqrt x$ and $y=x^2$ intersect at $(0,0$ and $(1,1)$.

Along $y=\sqrt x$, from $(0,0)$ to $(1,1)$ the line integral equals,

$$\int_0^1 [3x^2-8x] dx + \dfrac{4\sqrt x-6x\sqrt x}{2\sqrt x} dx $$

$$=\int_0^1[3x^2-8x+2-3x]dx=\int_0^1[3x^2-11x+2]dx = [x^3-5.5x^2+2x]_0^1=-2.5$$

Along $y=x^2$, from $(1,1)$ to $(0,0)$, the line integral equals,

$$\int_1^0 [3x^2-8x^4]dx + [8x^3-12x^4]dx = \int_1^0 [8x^3-20x^4+3x^2] dx = [x^3-4x^5+2x^4]_1^0 =1$$

The required integral = $-2.5+1=-1.5$

using, $$\int_c M dx + N dy = \iint_c \left[\dfrac{∂N}{∂x} - \dfrac{∂M}{∂y}\right] dx dy$$

...

we shall have,

$$\int_{\sqrt x}^{x^{2}} (10y)dx = [5y^2]_{\sqrt x}^{x^{2}} = 5x^4-5x $$

then

$$\int_{x=0}^1 [5x^4-5x] dx = [x^5 - 2.5x^2]_0^1 = -1.5$$

so that the theorem is verified.

Maybe my question would be; how to get the second value $\dfrac{5}{3}$.

 

[Orodruin]

Repeat the same thing, but keeping in mind that the boundary now consists of three curve segments rather than two.

 

[pasmith]

Parts {a} and {b} ask you to verify Green's Theorem for two different regions. You have only done the first.

(That you get -1.5 rather than 1.5 for part (a) suggests you have made an error. Recall that the line integral must be traversed counterclockwise. That means that the curve y = x^2 is traversed from x = 1 to x = 0 and then the curve x = y^2 is traversed from y = 1 to y = 0. It may be simpler to parametrize these as (t, t^2) and (t^2, t) respectively.

 

[chwala]

Parts {a} and {b} ask you to verify Green's Theorem for two different regions. You have only done the first.

(That you get -1.5 rather than 1.5 for part (a) suggests you have made an error. Recall that the line integral must be traversed counterclockwise. That means that the curve y = x^2 is traversed from x = 1 to x = 0 and then the curve x = y^2 is traversed from y = 1 to y = 0. It may be simpler to parametrize these as (t, t^2) and (t^2, t) respectively.

There is no error.

That is only but the textbook solution, my solution is also correct ...if you check on the Green's theorem...You'll note that one can use either direction as the path is a closed one.

Most importantly, is to apply the steps correctly...

Part (a) should be correct, ...I just used the opposite direction but strictly stuck to the concept as required..if I switch directions for the closed curve then I will realize the positive value.

Arrrggh part (b) should also be easy...I didn't read question properly...I had initially thought that part (a) had two value of solutions.

 

[Orodruin]

You'll note that one can use either direction as the path is a closed one.

This is incorrect.

 

[mathwonk]

if you believe wikipedia, the fact that the path integral should be taken in the counterclockwise direction, is stated there:
https://en.wikipedia.org/wiki/Green...een's theorem,special case of Stokes' theorem.

 

[chwala]

This is incorrect.

Noted, then my only option is to read on the concept again. Cheers.

 

[chwala]

if you believe wikipedia, the fact that the path integral should be taken in the counterclockwise direction, is stated there:
https://en.wikipedia.org/wiki/Green's_theorem#:~:text=In vector calculus, Green's theorem,special case of Stokes' theorem.

That is correct.

 

[chwala]

We shall then have,

Along $y=x^2$, from $(0,0)$ to $(1,1)$, the line integral equals,

$$\int_0^1 [3x^2-8x^4]dx + [8x^3-12x^4]dx = \int_0^1 [8x^3-20x^4+3x^2] dx = [x^3-4x^5+2x^4]_0^1 =-1$$


Along $y=\sqrt x$, from $(1,1)$ to $(0,0)$ the line integral equals,

$$\int_1^0 [3x^2-8x] dx + \dfrac{4\sqrt x-6x\sqrt x}{2\sqrt x} dx $$

$$=\int_1^0[3x^2-8x+2-3x]dx=\int_1^0[3x^2-11x+2]dx = [x^3-5.5x^2+2x]_1^0=2.5$$

The required line integral = $2.5-1=1.5$

...

also,

$$\int_{x^{2}}^{\sqrt x} (10y)dx = [5y^2]_{x^{2}}^{\sqrt x} = 5x-5x^4 $$

then

$$\int_{x=0}^1 [5x-5x^4] dx = [2.5x-x^5]_0^1 = 2.5-1=1.5$$

so that the theorem is verified.

 

[SammyS]

We shall then have,

Along $y=x^2$, from $(0,0)$ to $(1,1)$, the line integral equals,
$$\int_0^1 [3x^2-8x^4]dx + [8x^3-12x^4]dx = \int_0^1 [8x^3-20x^4+3x^2] dx = [x^3-4x^5+2x^4]_0^1 =-1$$
Along $y=\sqrt x$, from $(1,1)$ to $(0,0)$ the line integral equals,
$$\int_1^0 [3x^2-8x] dx + \dfrac{4\sqrt x-6x\sqrt x}{2\sqrt x} dx $$
$$=\int_1^0[3x^2-8x+2-3x]dx=\int_1^0[3x^2-11x+2]dx = [x^3-5.5x^2+2x]_1^0=2.5$$
The required line integral = $2.5-1=1.5$
...

also,
$$\int_{x^{2}}^{\sqrt x} (10y)dx = [5y^2]_{x^{2}}^{\sqrt x} = 5x-5x^4 $$
then
$$\int_{x=0}^1 [5x-5x^4] dx = [2.5x-x^5]_0^1 = 2.5-1=1.5$$
so that the theorem is verified.

Although you haven't stated what you're doing in this post or how it differs from the OP, it's apparent that here you have done the line integral in the counterclockwise (positive) sense.

You have also switched the sense in which you did the area (double) integral. - now going from the lower boundary to the upper boundary. In the OP, it was puzzling how you got a negative result for this integral, when the integrand was a positive quantity over the interior of the region of integration.

You still have an error in notation in the following.. Maybe it's simply a typo.

$$\int_{x^{2}}^{\sqrt x} (10y)dx = [5y^2]_{x^{2}}^{\sqrt x} = 5x-5x^4 $$

The integral, $\displaystyle \ \int_{x^{2}}^{\sqrt x} (10y)dx \$, is non-sense.

 

[chwala]

Although you haven't stated what you're doing in this post or how it differs from the OP, it's apparent that here you have done the line integral in the counterclockwise (positive) sense.

You have also switched the sense in which you did the area (double) integral. - now going from the lower boundary to the upper boundary. In the OP, it was puzzling how you got a negative result for this integral, when the integrand was a positive quantity over the interior of the region of integration.

You still have an error in notation in the following.. Maybe it's simply a typo.

The integral, $\displaystyle \ \int_{x^{2}}^{\sqrt x} (10y)dx \$, is non-sense.

typo...should be $\int_{x^{2}}^{\sqrt x} (10y)dy$ cheers.