Math Problem Statement

Use Green's theorem to evaluate int C log(1+y)dx - (xy)/(1 + y) * dy where with vertices (0,0),(2,0) and (0,4). Cis the triangle

Solution

Problem: Use Green's Theorem to evaluate the line integral:

Clog(1+y)dxxy1+ydy\int_C \log(1+y) \, dx - \frac{xy}{1+y} \, dy where CC is the boundary of the triangle with vertices (0,0)(0, 0), (2,0)(2, 0), and (0,4)(0, 4).


Green's Theorem:

Green's theorem states that for a curve CC positively oriented and enclosing a region RR: CMdx+Ndy=R(NxMy)dA\int_C M \, dx + N \, dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA Here:

  • M(x,y)=log(1+y)M(x, y) = \log(1 + y)
  • N(x,y)=xy1+yN(x, y) = -\frac{xy}{1 + y}

Step 1: Compute the partial derivatives

  1. Compute Nx\frac{\partial N}{\partial x}:
    N(x,y)=xy1+y,Nx=y1+yN(x, y) = -\frac{xy}{1 + y}, \quad \frac{\partial N}{\partial x} = -\frac{y}{1 + y}

  2. Compute My\frac{\partial M}{\partial y}:
    M(x,y)=log(1+y),My=11+yM(x, y) = \log(1 + y), \quad \frac{\partial M}{\partial y} = \frac{1}{1 + y}

  3. Compute NxMy\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}:
    NxMy=y1+y11+y=y+11+y=1\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -\frac{y}{1 + y} - \frac{1}{1 + y} = -\frac{y + 1}{1 + y} = -1


Step 2: Set up the double integral over RR

The region RR is the triangle with vertices (0,0)(0, 0), (2,0)(2, 0), and (0,4)(0, 4).
The limits of integration can be described as:

  • xx ranges from 00 to 22
  • For a fixed xx, yy ranges from 00 to 42x4 - 2x

Thus: R1dA=R1dA=02042x1dydx\iint_R -1 \, dA = -\iint_R 1 \, dA = -\int_0^2 \int_0^{4 - 2x} 1 \, dy \, dx


Step 3: Evaluate the double integral

  1. Compute the inner integral (yy-integration):
    042x1dy=[y]042x=(42x)0=42x\int_0^{4 - 2x} 1 \, dy = [y]_0^{4 - 2x} = (4 - 2x) - 0 = 4 - 2x

  2. Compute the outer integral (xx-integration):
    02(42x)dx=024dx022xdx\int_0^2 (4 - 2x) \, dx = \int_0^2 4 \, dx - \int_0^2 2x \, dx 024dx=4x02=8,022xdx=x202=4\int_0^2 4 \, dx = 4x \Big|_0^2 = 8, \quad \int_0^2 2x \, dx = x^2 \Big|_0^2 = 4 02(42x)dx=84=4\int_0^2 (4 - 2x) \, dx = 8 - 4 = 4

  3. Multiply by 1-1:
    R1dA=4-\iint_R 1 \, dA = -4


Final Answer:

Clog(1+y)dxxy1+ydy=4\int_C \log(1+y) \, dx - \frac{xy}{1+y} \, dy = -4


Do you have any questions or need clarification? 😊
Here are 5 related problems to explore:

  1. Apply Green’s theorem to evaluate a line integral where MM and NN are more complex functions.
  2. Determine the area of a triangle using Green’s theorem.
  3. Solve a similar problem where CC is a rectangle instead of a triangle.
  4. Verify Green's theorem numerically for a given function.
  5. Explore the conditions under which Green's theorem can be applied.

Tip: Always ensure the curve CC is oriented positively (counterclockwise) when using Green’s theorem!

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Green's Theorem
Line Integrals
Double Integrals

Formulas

Green's Theorem: ∫_C M dx + N dy = ∬_R (∂N/∂x - ∂M/∂y) dA
Partial derivatives: ∂N/∂x = -y/(1+y), ∂M/∂y = 1/(1+y)
Double integral setup: ∬_R 1 dA with appropriate limits of integration

Theorems

Green's Theorem

Suitable Grade Level

Undergraduate (Calculus II or III)