For a simple closed curve C in the plane show by Green's theorem that the area inclosed is A= ½ f (x dy – y dr). Use Problem 6 to show that the area inside the ellipse x = a cos0, y = bsin 0, 0 < θ < 2π, is A = nab.
For a simple closed curve C in the plane show by Green's theorem that the area inclosed is A= ½ f (x dy – y dr). Use Problem 6 to show that the area inside the ellipse x = a cos0, y = bsin 0, 0 < θ < 2π, is A = nab.
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Answer number 7
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![**6.** For a simple closed curve \( C \) in the plane, show by Green's theorem that the area enclosed is
\[
A = \frac{1}{2} \oint_C (x \, dy - y \, dx).
\]
**7.** Use Problem 6 to show that the area inside the ellipse \( x = a \cos \theta \), \( y = b \sin \theta \), \( 0 \leq \theta \leq 2\pi \), is \( A = \pi ab \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc47a605b-1167-4b41-bc40-883b330bf596%2Ff3797780-b564-46f5-8dd9-c2314ff527ef%2Frhw0q1k_processed.png&w=3840&q=75)
Transcribed Image Text:**6.** For a simple closed curve \( C \) in the plane, show by Green's theorem that the area enclosed is
\[
A = \frac{1}{2} \oint_C (x \, dy - y \, dx).
\]
**7.** Use Problem 6 to show that the area inside the ellipse \( x = a \cos \theta \), \( y = b \sin \theta \), \( 0 \leq \theta \leq 2\pi \), is \( A = \pi ab \).
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Here we will find that the area of ellipse is (pi)ab
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