Double Integration vs Line integration Question

TheSetMiner

Hi, I'm stuck on a question which involves green theorem and proving it by equating a line integral and a double integral. I have gotten the double integral to work out at 40 which I think is right but I can't follow the line integration at all. I guessed I'd need a parametrisation to work with so I let x = acost and y = asint but my answer was negative and not the same as my former answer.

Anyway, here is the question, I'd love is someone could point me in the right direction. Thanks!

---

By calculating both the line integral over the boundary and the double integral over the
area, verify Green’s theorem for the integral

Integral[c]((3x^2 − 8y^2)dx + (4y − 6xy)dy

where C is the boundary of the region enclosed by the line y = 0 and the curve y = 1−x^2

---

TheBody

Hi there.

The parameterisation you are using is incorrect. The one you are using is for a circle of radius a.

You will need to break up the integral around the boundary into two. Firstly, the bit alone the x-axis and then the curved bit.

A parameterisation for the bit along the x-axis could be:

[latex]C_1:[/latex] Let [latex]x=t[/latex] and [latex]y=0[/latex] where [latex]t=[-1,1][/latex] .

Then for the second integral, a parameterisation for the curved part could be:

[latex]C_2:[/latex] Let [latex]x=t[/latex] and [latex]y=1-t^2[/latex] where [latex]t=[-1,1][/latex] .


Give that a go and see how you get on.

TheSetMiner

TheBody wrote: »

Hi there.

The parameterisation you are using is incorrect. The one you are using is for a circle of radius a.

You will need to break up the integral around the boundary into two. Firstly, the bit alone the x-axis and then the curved bit.

A parameterisation for the bit along the x-axis could be:

[latex]C_1:[/latex] Let [latex]x=t[/latex] and [latex]y=0[/latex] where [latex]t=[-1,1][/latex] .

Then for the second integral, a parameterisation for the curved part could be:

[latex]C_2:[/latex] Let [latex]x=t[/latex] and [latex]y=1-t^2[/latex] where [latex]t=[-1,1][/latex] .


Give that a go and see how you get on.

Just revisiting this question and tried your method but my answers arent matching unfortunately.

for my double integral, my limits are [latex]y = 1-x^2, y = 0[/latex] and [latex]x=1, x=-1[/latex]
However, this integral is giving me 4/3 whereas your integral in terms of t altogether integrated between 1 and -1 gives me 22/3. I am watching youtube videos on the topic now and I am beginning to think the limits might be letting me down as I am not very confident in my selections.

TheSetMiner

the parameterisation is puzzling me also. Why do we need to look at the bit that is just along the x-axis, surely that area is zero? And does parametrisation merely involve subbing a variable t into the equation y(x) for x?

TheBody

TheSetMiner wrote: »

the parameterisation is puzzling me also. Why do we need to look at the bit that is just along the x-axis, surely that area is zero? And does parametrisation merely involve subbing a variable t into the equation y(x) for x?

Hi there.

I'm afraid I'm up to my neck at work today so I won't have time to go through the calculations for you.

Greens theorem gives us a way of turning an integration around a boundary into an integration over an area.

To do the boundary part, we need to parameterise the bit along the x-axis seperately to the curved bit as we cannot describe both at the same time.

TheSetMiner

TheBody wrote: »

Hi there.

I'm afraid I'm up to my neck at work today so I won't have time to go through the calculations for you.

Greens theorem gives us a way of turning an integration around a boundary integration over an area.

To do the boundary part, we need to parameterise the bit along the x-axis seperately to the curved bit as we cannot describe both at the same time.

No problem. thanks for your help