Use Green’s Theorem to evaluate the line integral along the given positively oriented curve.
∮
C
(ye
x
dx+2e
x
dy)
C is the rectangle with vertices (0, 0), (3, 0), (3, 2), and (0, 2).
The Correct Answer and Explanation is :
To evaluate the line integral
[
\oint_C \left( y e^x \, dx + 2 e^x \, dy \right)
]
using Green’s Theorem, we first identify the functions involved and set up the theorem correctly. Green’s Theorem states that if (C) is a positively oriented, simple, closed curve in the plane and (D) is the region bounded by (C), then:
[
\oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y} \right) dA
]
where (P = y e^x) and (Q = 2 e^x).
Step 1: Compute the Partial Derivatives
We need to compute the partial derivatives of (P) and (Q):
- (P = y e^x)
[
\frac{\partial P}{\partial y} = e^x
]
- (Q = 2 e^x)
[
\frac{\partial Q}{\partial x} = 2 e^x
]
Step 2: Apply Green’s Theorem
Now we can substitute these into Green’s Theorem:
[
\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y} = 2 e^x – e^x = e^x
]
Step 3: Set Up the Double Integral
Next, we need to evaluate the double integral over the region (D), which is the rectangle with vertices ((0, 0)), ((3, 0)), ((3, 2)), and ((0, 2)):
[
\iint_D e^x \, dA = \int_0^3 \int_0^2 e^x \, dy \, dx
]
The inner integral with respect to (y) is straightforward:
[
\int_0^2 e^x \, dy = e^x \cdot y \bigg|_0^2 = 2 e^x
]
Step 4: Evaluate the Outer Integral
Now we compute the outer integral:
[
\int_0^3 2 e^x \, dx = 2 \left( e^x \bigg|_0^3 \right) = 2 (e^3 – e^0) = 2 (e^3 – 1)
]
Conclusion
Thus, the value of the line integral is:
[
\oint_C \left( y e^x \, dx + 2 e^x \, dy \right) = 2 (e^3 – 1)
]
Green’s Theorem allows us to convert a complex line integral into a more manageable double integral, illustrating the beauty of calculus in connecting line and area calculations.