Suppose we want to calculate the area using green theorem. First, we want to calculate the area of the colored part of the image attached. We can calculate that using double integral or line integral of the field F = (0,x) Calculating using the field, we have to know the parametrization, which is r(t) = (sin(t), sin(t)cos(t)), 0 <= t <= pi We know the limits, we know the field, we just have to calculate the term dr. Then, we need to calculate the integral: see 2n image.
Suppose we want to calculate the area using green theorem. First, we want to calculate the area of the colored part of the image attached. We can calculate that using double
Calculating using the field, we have to know the parametrization, which is r(t) = (sin(t), sin(t)cos(t)), 0 <= t <= pi
We know the limits, we know the field, we just have to calculate the term dr. Then, we need to calculate the integral: see 2n image.
The final answer is 2/3 and its fine.
But, suppose we have another question to calculate the area by using green theorem. At this time, the region is determined by the pair of curves:
y = x3 - x and y = x2 - 1
I know we have to calculate the limits, and that they are -1 and 1.
We could calculate using the same field F(0,x)
Then we have to parametrize the equations and then calculate the integral of xdy.
The thing is at this time we have to calculate two integrals ∫cF2dy = ∫c1xdy + ∫c2xdy and sum them, each one is related to one curve.
Why is that?
Please explain why in the first one we have to calculate just one integral, and in the second one we need to calculate two integrals. No need to calculate them.
![-
F. dr = *(0, sin(t)) - (cos(t), cos²(t) — sin²(t))dt =
C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb04a77a-4ed3-4dfc-a5cd-1224d7e51a50%2Fd30d83fe-7c79-47a3-9055-bddd2aec80ff%2Fhtb85xv_processed.jpeg&w=3840&q=75)
![X
-0.5
-0.4
0.5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb04a77a-4ed3-4dfc-a5cd-1224d7e51a50%2Fd30d83fe-7c79-47a3-9055-bddd2aec80ff%2F1mgjr3o_processed.jpeg&w=3840&q=75)
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