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Math  /  Calculus

Question3. Use Green's Theorem to evaluate ∮C2xydx+(x+y)dy\oint_{C} 2 x y d x+(x+y) d y. CC is the boundary of the region lying between the graphs of y=0y=0 and y=1−x2y=1-x^{2}, oriented counterclockwise.

Studdy Solution

STEP 1

1. Green's Theorem relates a line integral around a simple closed curve C C to a double integral over the plane region D D bounded by C C .
2. The vector field F=⟨2xy,x+y⟩\mathbf{F} = \langle 2xy, x+y \rangle is continuously differentiable over the region D D .
3. The region D D is the area between y=0 y = 0 and y=1−x2 y = 1 - x^2 .

STEP 2

1. Verify the applicability of Green's Theorem.
2. Set up the double integral using Green's Theorem.
3. Evaluate the double integral over the region D D .

STEP 3

Verify that Green's Theorem can be applied. The theorem states:
∮CF⋅dr=∬D(∂N∂x−∂M∂y)dA\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA
where F=⟨M,N⟩=⟨2xy,x+y⟩\mathbf{F} = \langle M, N \rangle = \langle 2xy, x+y \rangle.
Check the partial derivatives: - ∂N∂x=∂∂x(x+y)=1\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x+y) = 1 - ∂M∂y=∂∂y(2xy)=2x\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2xy) = 2x
Green's Theorem is applicable since the region D D is simply connected and the vector field is continuously differentiable.

STEP 4

Set up the double integral over the region D D . The region D D is bounded by y=0 y = 0 and y=1−x2 y = 1 - x^2 .
The double integral becomes:
∬D(1−2x)dA\iint_{D} \left( 1 - 2x \right) dA
Express the region D D in terms of x x and y y : - The limits for y y are from 0 0 to 1−x2 1 - x^2 . - The limits for x x are from −1 -1 to 1 1 (since y=1−x2 y = 1 - x^2 is a downward-opening parabola intersecting the x-axis at x=±1 x = \pm 1 ).
The double integral becomes:
∫−11∫01−x2(1−2x) dy dx\int_{-1}^{1} \int_{0}^{1-x^2} (1 - 2x) \, dy \, dx

STEP 5

Evaluate the double integral:
First, integrate with respect to y y :
∫01−x2(1−2x) dy=(1−2x)⋅y∣01−x2=(1−2x)(1−x2)\int_{0}^{1-x^2} (1 - 2x) \, dy = (1 - 2x) \cdot y \bigg|_{0}^{1-x^2} = (1 - 2x)(1-x^2)
Now, integrate with respect to x x :
∫−11(1−2x)(1−x2) dx\int_{-1}^{1} (1 - 2x)(1-x^2) \, dx
Expand the integrand:
=∫−11(1−x2−2x+2x3) dx= \int_{-1}^{1} (1 - x^2 - 2x + 2x^3) \, dx
Separate the integral:
=∫−111 dx−∫−11x2 dx−∫−112x dx+∫−112x3 dx= \int_{-1}^{1} 1 \, dx - \int_{-1}^{1} x^2 \, dx - \int_{-1}^{1} 2x \, dx + \int_{-1}^{1} 2x^3 \, dx
Evaluate each integral:
1. ∫−111 dx=[x]−11=1−(−1)=2\int_{-1}^{1} 1 \, dx = [x]_{-1}^{1} = 1 - (-1) = 2
2. ∫−11x2 dx=[x33]−11=13−(−13)=23\int_{-1}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_{-1}^{1} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}
3. ∫−112x dx=[x2]−11=1−1=0\int_{-1}^{1} 2x \, dx = [x^2]_{-1}^{1} = 1 - 1 = 0
4. ∫−112x3 dx=[x42]−11=12−12=0\int_{-1}^{1} 2x^3 \, dx = \left[\frac{x^4}{2}\right]_{-1}^{1} = \frac{1}{2} - \frac{1}{2} = 0
Combine the results:
2−23+0+0=63−23=432 - \frac{2}{3} + 0 + 0 = \frac{6}{3} - \frac{2}{3} = \frac{4}{3}
The value of the line integral is:
43\boxed{\frac{4}{3}}

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