Use Green's theorem to evaluate the integral:
y^(2)dx+xy dy
where C is the boundary of the region lying between the graphs of y=0,
y=sqrt(x), and x=9
P = y^2
Q = xy
�ç[0,9]�ç[0(�ãx)](�ÝQ/�Ýx-�Ýp/�Ýy)dy dx
�ç[0, 9]�ç[0, �ãx ](y-2y)dy dx
�ç[0,9]�ç[0, �ãx](-y)dy dx
�ç[0,9]-x/2 dx
=(-x^2)/4 [0, 9] = (-81)/4
To evaluate the integral using Green's theorem, we need to find the line integral of the given vector field over the boundary of the region.
First, let's find the partial derivatives of the vector field:
∂/∂x (y^2) = 0
∂/∂y (xy) = x
Now, we can write the line integral using Green's theorem:
∮C (y^2 dx + xy dy) = ∬R (∂/∂x (xy) - ∂/∂y (y^2)) dA
Where R is the region between the graphs y=0, y=sqrt(x), and x=9.
To evaluate the double integral, we need to find the limits of integration. The region is bounded by y=0, y=sqrt(x), and x=9. So the limits of integration for y are 0 to sqrt(x), and for x, it is from 0 to 9.
∬R (∂/∂x (xy) - ∂/∂y (y^2)) dA = ∫(0 to 9) ∫(0 to sqrt(x)) (x - 2y) dy dx
Now, let's integrate with respect to y first:
∫(0 to sqrt(x)) (x - 2y) dy = xy - y^2 |_(0 to sqrt(x))
= x(sqrt(x)) - (sqrt(x))^2 - (x(0) - (0)^2)
= x√x - x - 0
= x√x - x
Now, we integrate with respect to x:
∫(0 to 9) (x√x - x) dx = ∫(0 to 9) (x^(3/2) - x) dx
= (2/5)x^(5/2) - (1/2)x^2 |_(0 to 9)
= (2/5)(9)^(5/2) - (1/2)(9)^2 - (2/5)(0)^(5/2) + (1/2)(0)^2
= (2/5)(9)^(5/2) - (1/2)(9)^2
Therefore, the value of the given integral using Green's theorem is (2/5)(9)^(5/2) - (1/2)(9)^2.
To use Green's theorem to evaluate the given integral, we need to first determine the orientation of the boundary curve C.
The boundary C consists of three parts:
1) The curve y=0 for 0≤x≤9, which lies along the x-axis.
2) The curve y=√x for 0≤x≤9, which is a portion of the parabolic curve y²=x.
3) The line x=9, which is a vertical line.
To determine the orientation, we need to travel along the boundary C in a counterclockwise direction. Let's break down the integral into three separate integrals corresponding to each part of the boundary curve.
1) Along the curve y=0 for 0≤x≤9:
Parameterize the curve by setting y=0 and letting x vary from 0 to 9:
r₁(t) = ⟨t, 0⟩, where 0≤t≤9
dr₁ = ⟨dt, 0⟩
2) Along the curve y=√x, for 0≤x≤9:
Parameterize the curve by letting y=√x and x vary from 0 to 9:
r₂(t) = ⟨t, √t⟩, where 0≤t≤9
dr₂ = ⟨dt, (1/2√t)dt⟩
3) Along the line x=9:
Parameterize the line by letting x=9 and y vary from 0 to √9=3:
r₃(t) = ⟨9, t⟩, where 0≤t≤3
dr₃ = ⟨0, dt⟩
Now, we can evaluate the three separate integrals:
1) ∫ᶜ (y²dx + xdy) along the curve y=0 for 0≤x≤9:
∫₀⁹ (0²dx + x(0dy))
= ∫₀⁹ (0 dx)
= 0
2) ∫ᶜ (y²dx + xdy) along the curve y=√x for 0≤x≤9:
∫₀⁹ (√x²dx + x(d√x))
= ∫₀⁹ (x dx + x(1/2√x) dx)
= ∫₀⁹ (3/2x dx)
= (3/2) [x²/2]₀⁹
= (3/2)(81/2 - 0)
= 81/4
3) ∫ᶜ (y²dx + xdy) along the line x=9:
∫₀³ (y²(0)dy + 9dy)
= ∫₀³ (9dy)
= 9[ y ]₀³
= 9(3 - 0)
= 27
Finally, we sum up the three separate integrals:
0 + 81/4 + 27 = 81/4 + 27 = 243/4
Therefore, the value of the integral ∫ᶜ (y²dx + xdy) along the boundary C is 243/4.