Use Green's Theorem to evaluate
C (F · dr)
(Check the orientation of the curve before applying the theorem.)
F(x, y) = e^(2x) + x^(2)y, e^(2y) − xy^2
C is the circle x^2 + y^2 = 4 oriented clockwise
Do you mean
F(x,y) = e^(2x) + x^(2)yi + e^(2y) − xy^2 j ?
If so, then
F•dr = (e^(2x) + x^2y) dx + (e^(2y) − xy^2) dy
That makes
My = x^2
Nx = -y^2
∫F.dr = -∫∫x^2+y^2 dy dx
Using polar coordinates, that is just
-∫∫r^2 * r dr dθ
inside the circle:
r = [0,2]
θ = [0,2pi]
To evaluate the line integral C (F · dr) using Green's theorem, we need to express the line integral as a double integral over the region enclosed by the curve C.
First, let's check the orientation of the curve:
C is the circle x^2 + y^2 = 4 oriented clockwise.
Since C is oriented clockwise, we need to negate the expression for C (F · dr) before applying Green's theorem.
Now, let's calculate the double integral over the region enclosed by C using Green's theorem:
Green's theorem states that for a vector field F = (P, Q) and a curve C oriented positively (counterclockwise) enclosing a region D in the xy-plane, the line integral C (F · dr) can be evaluated as the double integral over the region D of the curl of F, which is given by:
∬(curl F) dA = ∬(∂Q/∂x - ∂P/∂y) dA,
where dA represents the differential area element in the xy-plane.
Let's find the curl of F:
curl F = ∂Q/∂x - ∂P/∂y
= ∂(e^(2y) − xy^2)/∂x - ∂(e^(2x) + x^(2)y)/∂y
= -2x - 2x
= -4x.
Now, let's write the double integral using the curl of F:
∬(curl F) dA = ∬(-4x) dA.
Since the curve C is a circle, the region D enclosed by C is the disk with radius 2 centered at the origin. We can express the double integral as an iterated integral over this disk:
∬(-4x) dA = ∫[0, 2π] ∫[0, 2] (-4rcosθ) r dr dθ,
where r is the radial distance from the origin and θ is the angle measured counterclockwise from the positive x-axis.
Evaluating the inner integral with respect to r, we get:
∫[0, 2π] ∫[0, 2] (-4rcosθ) r dr dθ
= ∫[0, 2π] [-2r^3cosθ]∣[0, 2] dθ
= ∫[0, 2π] (-16cosθ) dθ
= -16∫[0, 2π] cosθ dθ
= -16(sinθ)∣[0, 2π]
= 0.
Therefore, the value of the line integral C (F · dr) using Green's theorem is 0.
To evaluate the line integral C (F · dr) using Green's Theorem, we need to follow these steps:
Step 1: Check the orientation of the curve C.
In this case, C is a circle given by the equation x^2 + y^2 = 4, and it is oriented clockwise. This means that the curve will be traced in the opposite direction of the positive orientation (counterclockwise).
Step 2: Express the vector field F as F = (M, N).
For F(x, y) = e^(2x) + x^(2)y, e^(2y) - xy^2, we have M(x, y) = e^(2x) + x^2y and N(x, y) = e^(2y) - xy^2.
Step 3: Compute the partial derivatives of M and N with respect to x and y.
∂M/∂x = 2e^(2x) + 2xy
∂N/∂y = 2e^(2y) - 2xy
Step 4: Apply Green's Theorem.
According to Green's Theorem, the line integral C (F · dr) can be expressed as the double integral of ( ∂N/∂x - ∂M/∂y ) over the region R enclosed by the curve C.
∫∫( ∂N/∂x - ∂M/∂y ) dA
Step 5: Evaluate the double integral over the region R.
Since the curve C is a circle, we can use polar coordinates to parametrize the region R.
Let r represent the distance from the origin and θ represent the angle. We have:
x = r cos(θ)
y = r sin(θ)
The equation of the circle x^2 + y^2 = 4 becomes r^2 = 4, or r = 2.
The region R can be represented as 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 2.
Now, substitute these parametric expressions into the equation from Step 4 and evaluate the double integral.
∫∫( ∂N/∂x - ∂M/∂y ) r dr dθ
Step 6: Simplify and calculate the integral.
First, compute the partial derivatives ∂N/∂x and ∂M/∂y using the expressions for M and N.
∂N/∂x = 2e^(2y)
∂M/∂y = x^2 + e^(2y)
Substituting these values into the equation, we get:
∫∫( 2e^(2y) - (x^2 + e^(2y)) ) r dr dθ
Now, solve this double integral by integrating r first and then integrating θ.
∫0 to 2π ∫0 to 2 ( 2e^(2y) - (x^2 + e^(2y)) ) r dr dθ
Evaluate the inner integral with respect to r:
∫0 to 2π ( e^(2y)r^2 - (1/3)x^3 - e^(2y)r ) |0 to 2 dθ
Evaluate each term at the limits of integration and simplify:
∫0 to 2π (e^(2y)(4) - (1/3)(8) - e^(2y)(2)) dθ
Simplify further:
∫0 to 2π (4e^(2y) - (8/3) - 2e^(2y)) dθ
Combine like terms:
∫0 to 2π (2e^(2y) - (8/3)) dθ
Now, integrate with respect to θ:
[2e^(2y)θ - (8/3)θ] from 0 to 2π
Evaluate the expression at the limits of integration:
[2e^(2y)(2π) - (8/3)(2π)] - [2e^(2y)(0) - (8/3)(0)]
Simplify:
(4πe^(2y) - (16/3)π) - (0 - 0)
Finally, we have our solution:
(4πe^(2y) - (16/3)π) cubic units.