Answer: ∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ 5dA = 5/3
Step-by-step explanation:
Given that;
∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy)
Green's Theorem is given as;
∫c (P(x,y)dx + Q(x,y)dy) = ∫∫ₐ { (-β/βy) P(x,y) + (β/βy) Q(x,y) } dA
Now our P(x,y) = 3y + 7e^(√x) and our Q(x,y) = 8x + 5 cos (y²)
Since we know this, therefore; we substitute
∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ { (-β/βy) (3y + 7e^(√x)) + (β/βy) (8x + 5 cos (y²)) } dA
∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ ( 8-3) dA
∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ 5dA
from the question, our region is defined by a lower bound: y = x² and an upper bound of y = √x
going from x = 0 to x = 1
Now calculating ∫∫ₐ 5dA by means of the description of the region, we say;
∫∫ₐ 5dA = 5¹∫₀ ₓ²∫^(√x) dydx
∫∫ₐ 5dA = 5¹∫₀ (y)∧(y-√x) ∨(y-x²) dx
∫∫ₐ 5dA = 5¹∫₀ (√x-x²) dx
∫∫ₐ 5dA = 5 [ ((x^(3/2))/(3/2)) - x³/3]¹₀ NOW since ∫[f(x)]ⁿ dx = ([f(x)]ⁿ⁺¹ / n+1) + C
then
∫∫ₐ 5dA = 5 [ ((1^(3/2))/(3/2)) - 1³ / 3) - ((0^(3/2))/(3/2)) - 0³ / 3) ]
∫∫ₐ 5dA = 5 [ ((1^(3/2))/(3/2)) - 1³ / 3)
∫∫ₐ 5dA = 5/3
Therefore ∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ 5dA = 5/3
