Use Green's Theorem to evaluate the integral In(x2 + 1)dx – xdy, C is the boundary of the region bounded by y=V4- x2 and y=0 in a counterclock wise direction - 21 а. b. с. d. е. - 3 T
Arc Length
Arc length can be thought of as the distance you would travel if you walked along the path of a curve. Arc length is used in a wide range of real applications. We might be interested in knowing how far a rocket travels if it is launched along a parabolic path. Alternatively, if a curve on a map represents a road, we might want to know how far we need to drive to get to our destination. The distance between two points along a curve is known as arc length.
Line Integral
A line integral is one of the important topics that are discussed in the calculus syllabus. When we have a function that we want to integrate, and we evaluate the function alongside a curve, we define it as a line integral. Evaluation of a function along a curve is very important in mathematics. Usually, by a line integral, we compute the area of the function along the curve. This integral is also known as curvilinear, curve, or path integral in short. If line integrals are to be calculated in the complex plane, then the term contour integral can be used as well.
Triple Integral
Examples:
![### Use Green's Theorem to Evaluate the Integral
Evaluate the following line integral using Green's Theorem:
\[ \oint_{C} \ln(x^2+1)dx - xdy \]
where \(C\) is the boundary of the region bounded by \( y = \sqrt{4 - x^2} \) and \( y = 0 \) in a counterclockwise direction.
### Multiple Choice Options:
a. \(-2\pi\)
b. \(\pi\)
c. \(3\pi\)
d. \(4\pi\)
e. \(-3\pi\)
#### Explanation of Terms:
- **\[ \oint_{C} \]** represents a line integral around the closed curve \(C\).
- **\[ \ln(x^2 + 1)dx - xdy \]** is the given integrand.
- **\( C \)** is the boundary described by the parabola \( y = \sqrt{4 - x^2} \) and the line \( y = 0 \).
#### Concepts Involved:
1. **Green's Theorem:** It relates the line integral around a simple closed curve \(C\) to a double integral over the plane region \(D\) bounded by \(C\).
\[ \oint_{C} (Ldx + Mdy) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA \]
In this problem:
- \( L(x, y) = \ln(x^2 + 1) \)
- \( M(x, y) = -x \)
2. **Region \(D\):**
- The region \(D\) is bounded by the curve \( y = \sqrt{4 - x^2} \) and the line \( y = 0 \).
- This corresponds to the upper semicircle of radius 2 centered at the origin, extending from \( x = -2 \) to \( x = 2 \).
3. **Counterclockwise Direction:** This implies the orientation of the integration path is such that the region \(D\) is always on the left side as you traverse the curve \(C\).
By implementing Green's Theorem, the given line integral can be converted into a domain integral over \(D\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee1f5d00-2b06-4d9b-996b-2e4c16434e95%2F4a73ecca-b4b0-481a-9253-b924f21befe9%2F8ssmgnt_processed.jpeg&w=3840&q=75)
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