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
To use Green's theorem to evaluate the given integral, we first need to find the curl of the vector field.
The vector field is given as F = (y^2, xy).
The curl of a vector field F = (P, Q) in two dimensions is defined as:
curl(F) = (∂Q/∂x) - (∂P/∂y)
Let's find the partial derivatives of Q = xy:
∂Q/∂x = y
And the partial derivatives of P = y^2:
∂P/∂y = 2y
Now, we can calculate the curl:
curl(F) = (∂Q/∂x) - (∂P/∂y) = y - 2y = -y
Now that we have the curl, we can use Green's theorem to evaluate the given integral.
Green's theorem states that for a simply-connected region R bounded by a positively oriented, piecewise-smooth, simple closed curve C, the integral of the curl of a vector field F over C is equal to the double integral of F over the region R:
∮C F · dr = ∬R curl(F) dA
In this case, F = (y^2, xy), curl(F) = -y, and the region R is bounded by the curves y = 0, y = √x, and x = 9.
Now, we need to find the limits of integration for the double integral. For this region, we can choose x as the outer integral and y as the inner integral. The limits for x are from 0 to 9, and for y, it goes from 0 to √x.
So, the integral becomes:
∫[0,9] ∫[0,√x] -y dy dx
We can solve this integral by following these steps:
1. Solve the inner integral with respect to y:
∫[0,√x] -y dy = [-y^2/2] [0,√x] = -√x^2/2 + 0/2 = -√x^2/2
2. Now, substitute the result back into the original integral and solve the outer integral:
∫[0,9] -√x^2/2 dx = [-x^3/6] [0,9] = -9^3/6 - 0/6 = -729/6
Therefore, the value of the given integral using Green's theorem is -729/6.