Using Green's Theorem evaluate the integral ∮c(xydx + x^2y^2 dy) where C is the triangle with vertices (0 ,0), (1, 0) and (1, 2).
well, you have
M = xy
N = x^2y^2
so the integral is just
∫∫R (Nx - My) dx dy
Along y=0, dy=0,
Along x=1, dx=0,
Along y=2x, dy = 2dx
Now just break the boundary into three parts, and evaluate each part. Watch the direction of the limits.
To evaluate the given line integral using Green's theorem, we need to find the curl of the vector field and then compute the line integral by converting it into a double integral over the region enclosed by the curve.
Step 1: Find the curl of the vector field F = (xy, x^2y^2):
The curl of a vector field F = (M, N) in two dimensions is defined as the partial derivative of N with respect to x minus the partial derivative of M with respect to y.
Taking the partial derivatives, we have:
∂M/∂y = x
∂N/∂x = 2xy^2
Therefore, the curl of F is given by:
curl(F) = ∂N/∂x - ∂M/∂y = 2xy^2 - x
Step 2: Apply Green's theorem to convert the line integral into a double integral:
Green's theorem states that the line integral of a vector field F around a simple closed curve C is equal to the double integral of the curl of F over the region D enclosed by C.
∮c(xydx + x^2y^2 dy) is equal to the double integral of curl(F) over D.
In this case, C is a triangle with vertices (0, 0), (1, 0), and (1, 2), which encloses the region D.
Step 3: Determine the limits of integration for the double integral:
The triangle C can be expressed as a region D with the following inequalities:
0 ≤ x ≤ 1
0 ≤ y ≤ 2x
So the limits of integration are:
∫[0 to 1] ∫[0 to 2x] (2xy^2 - x) dy dx
Step 4: Evaluate the double integral:
∫[0 to 1] ∫[0 to 2x] (2xy^2 - x) dy dx
= ∫[0 to 1] [2xy^3/3 - xy] |[0 to 2x] dx
= ∫[0 to 1] (16x^4/3 - 2x^2) dx
Evaluating the integral gives the final answer.
To evaluate the integral ∮c(xydx + x^2y^2 dy) using Green's Theorem, we can start by finding the curl of the vector field F = (P, Q) where P = xy and Q = x^2y^2. The curl of a vector field is given by the partial derivatives of its components:
∂Q/∂x - ∂P/∂y
Let's compute these partial derivatives:
∂Q/∂x = ∂/∂x (x^2y^2) = 2xy^2
∂P/∂y = ∂/∂y (xy) = x
Now, substituting these values, we can find the curl of the vector field:
curl(F) = (2xy^2 - x)
Next, we apply Green's Theorem, which states that for a simply connected, piecewise smooth region R bounded by a positively oriented, piecewise smooth, simple closed curve C, the line integral of a vector field F over C is equal to the double integral of the curl of F over R:
∮c(xydx + x^2y^2 dy) = ∬R (2xy^2 - x) dA
Since the given region C is a triangle with vertices (0, 0), (1, 0), and (1, 2), we can express this region R as the area enclosed by these vertices. To find the limits of integration for the double integral, we use the extreme values of x and y within the region:
∬R (2xy^2 - x) dA = ∫[0,1] ∫[0,2x] (2xy^2 - x) dy dx
Now, let's evaluate this double integral. First, integrate with respect to y:
∫[0,2x] (2xy^2 - x) dy = xy^3 - xy | from 0 to 2x
= 2x(8x^3) - x(2x) - 0
= 16x^4 - 2x^2
Now, integrate the result with respect to x:
∫[0,1] (16x^4 - 2x^2) dx = (16/5)x^5 - (2/3)x^3 | from 0 to 1
= (16/5) - (2/3)
= 18/15 - 10/15
= 8/15
Therefore, the value of the line integral ∮c(xydx + x^2y^2 dy) over the given triangle C is 8/15.