Chapter 6: Q7P (page 314)

Use Problem 6 to show that the area inside the ellipse $x = a c o s θ, y = b s i n θ, 0 \leq θ \leq 2 π, i s A = π a b .$

Short Answer

Expert verified

The solution to this problem is $A = a b π .$ .

Step by step solution

Given Information.

The given information is as follows:

$x = a \cos θ, y = b \sin θ .$

Definition of Green’s Theorem.

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

Find the solution.

Write the given conditions.

$x = a \cos θ, y = b \sin θ .$

Write the differentiation.

$d x = - a \sin θ d θ d y = b \cos θ d θ$

Find the area.

$A = \frac{1}{2} \int_{0}^{2 π} a b (\cos^{2} θ + \sin^{2} θ) d θ = \frac{2 π a b}{2} = π a b$

Hence, the solution to this problem is role="math" localid="1659154628756" $A = a b π$

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Use Problem 6 to show that the area inside the ellipse $x = a c o s θ, y = b s i n θ, 0 \leq θ \leq 2 π, i s A = π a b .$

Short Answer

Step by step solution

Given Information.

Definition of Green’s Theorem.

Find the solution.

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Use Problem 6 to show that the area inside the ellipse x=acosθ,y=bsinθ,0≤θ≤2π,isA=πab.

Short Answer

Step by step solution

Given Information.

Definition of Green’s Theorem.

Find the solution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Measurements

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Study anywhere. Anytime. Across all devices.

Use Problem 6 to show that the area inside the ellipse $x = a c o s θ, y = b s i n θ, 0 \leq θ \leq 2 π, i s A = π a b .$