Math Problem Statement

Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integral ∫C 9y^2x dx + 4x^2y dy.

Solution

To evaluate the line integral C(9y2xdx+4x2ydy)\int_C (9y^2x \, dx + 4x^2y \, dy) using Green's Theorem, let's break it into steps.

Step 1: Green's Theorem

Green's Theorem states: C(Pdx+Qdy)=R(QxPy)dA\int_C (P \, dx + Q \, dy) = \int\int_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \, dA where:

  • CC is the positively oriented boundary of region RR,
  • P=9y2xP = 9y^2x,
  • Q=4x2yQ = 4x^2y.

Step 2: Compute Partial Derivatives

We compute:

  • Qx=x(4x2y)=8xy\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(4x^2y) = 8xy,
  • Py=y(9y2x)=18yx\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(9y^2x) = 18yx.

Thus: QxPy=8xy18xy=10xy.\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 8xy - 18xy = -10xy.

Step 3: Set Up the Double Integral

The region RR is the square with vertices (0,0),(1,0),(1,1),(0,1)(0, 0), (1, 0), (1, 1), (0, 1). For this square:

  • x[0,1]x \in [0, 1],
  • y[0,1]y \in [0, 1].

The double integral becomes: R(10xy)dA=0101(10xy)dxdy.\int\int_R (-10xy) \, dA = \int_0^1 \int_0^1 (-10xy) \, dx \, dy.

Step 4: Evaluate the Double Integral

First, integrate with respect to xx: 01(10xy)dx=0110yxdx=10y[x22]01=10y12=5y.\int_0^1 (-10xy) \, dx = \int_0^1 -10y \cdot x \, dx = -10y \left[\frac{x^2}{2}\right]_0^1 = -10y \cdot \frac{1}{2} = -5y.

Now, integrate with respect to yy: 01(5y)dy=015ydy=5[y22]01=512=52.\int_0^1 (-5y) \, dy = \int_0^1 -5 \cdot y \, dy = -5 \left[\frac{y^2}{2}\right]_0^1 = -5 \cdot \frac{1}{2} = -\frac{5}{2}.

Final Answer:

The value of the line integral is: C(9y2xdx+4x2ydy)=52.\int_C (9y^2x \, dx + 4x^2y \, dy) = -\frac{5}{2}.

Would you like additional details or an explanation for any part of this? Here are related questions to consider:

  1. What does Green's Theorem imply geometrically?
  2. How does the orientation of the curve affect the result?
  3. Could this problem be evaluated without Green's Theorem? How?
  4. How would the result change if the region RR were a triangle instead of a square?
  5. What role do partial derivatives play in Green's Theorem?

Tip: Always confirm the orientation (positive or negative) of the curve when applying Green’s Theorem!

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Math Problem Analysis

Mathematical Concepts

Line Integrals
Double Integrals
Partial Derivatives

Formulas

Green's Theorem: ∫C (P dx + Q dy) = ∬R (∂Q/∂x - ∂P/∂y) dA

Theorems

Green's Theorem

Suitable Grade Level

Undergraduate Calculus