x F. dr, where F(x, y) = ( Y y+3 x 4, 4).0 C is the path along (x − 5)² + (y + 3)² = 4, traversed counter-clockwise exactly once, starting at (7, -3) and ending at (7,-3). Determine whether the circulation form of Green's Theorem can be directly applied to evaluate this integral. " Consider the line integral F Select the correct answer below: Green's Theorem does not apply because the path C' is not closed. Green's Theorem does not apply because F(x, y) does not have continuous partials in the interior of C. O Green's Theorem can be directly applied to evaluate this integral.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Question
Consider the line integral 

\[
\int_{C} \mathbf{F} \cdot d\mathbf{r},
\]

where \(\mathbf{F}(x, y) = \left\langle \frac{x}{y + 3}, \frac{y}{x - 4} \right\rangle\). \(C\) is the path along \((x - 5)^2 + (y + 3)^2 = 4\), traversed counter-clockwise exactly once, starting at \((7, -3)\) and ending at \((7, -3)\). Determine whether the circulation form of Green's Theorem can be directly applied to evaluate this integral.

Select the correct answer below:
- \( \bigcirc \) Green’s Theorem does not apply because the path \(C\) is not closed.
- \( \bigcirc \) Green’s Theorem does not apply because \(\mathbf{F}(x, y)\) does not have continuous partials in the interior of \(C\).
- \( \bigcirc \) Green’s Theorem can be directly applied to evaluate this integral.
Transcribed Image Text:Consider the line integral \[ \int_{C} \mathbf{F} \cdot d\mathbf{r}, \] where \(\mathbf{F}(x, y) = \left\langle \frac{x}{y + 3}, \frac{y}{x - 4} \right\rangle\). \(C\) is the path along \((x - 5)^2 + (y + 3)^2 = 4\), traversed counter-clockwise exactly once, starting at \((7, -3)\) and ending at \((7, -3)\). Determine whether the circulation form of Green's Theorem can be directly applied to evaluate this integral. Select the correct answer below: - \( \bigcirc \) Green’s Theorem does not apply because the path \(C\) is not closed. - \( \bigcirc \) Green’s Theorem does not apply because \(\mathbf{F}(x, y)\) does not have continuous partials in the interior of \(C\). - \( \bigcirc \) Green’s Theorem can be directly applied to evaluate this integral.
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