Poll Results
No votes. Be the first one to vote.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
To use Green’s theorem to calculate the value for the given functions (F = y^2) and (G = x^2) across a specified region, we first need to understand the theorem in the context of a region (R) and its positively oriented boundary (C). The theorem states:
[oint_C (L dx + M dy) = int int_R left(frac{partial M}{partial x} – frac{partial L}{partial y}right) dA]
where (L) and (M) are the components of a vector field, that is, (mathbf{F} = Lmathbf{i} + Mmathbf{j}).
For the given functions, if we interpret (F = y^2) as (L) and (G = x^2) as (M), then we have:
– (L = F = y^2)
– (M = G = x^2)
To apply Green’s theorem, we need to evaluate (frac{partial M}{partial x} – frac{partial L}{partial y}):
[
frac{partial M}{partial x} = frac{partial (x^2)}{partial x} = 2x
]
[
frac{partial L}{partial y} = frac{partial (y^2)}{partial y} =
c
Explanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(2x – 2y)dx dy. On integrating for x = 0->1 and y
= 0->2, we get Green’s value as -2.