## Use the parametric equations of an ellipse, x=acosθ, y=bsinθ, 0≤θ≤2π , to find the area that it encloses.

Question

Use the parametric equations of an ellipse, x=acosθ, y=bsinθ, 0≤θ≤2π , to find the area that it encloses.

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1 year 2021-07-23T11:38:15+00:00 1 Answers 500 views 0

Area of ellipse=$$\pi ab$$

Step-by-step explanation:

We are given that

$$x=acos\theta$$

$$y=bsin\theta$$

$$0\leq\theta\leq 2\pi$$

We have to find the area enclose by it.

$$x/a=cos\theta, y/b=sin\theta$$

$$sin^2\theta+cos^2\theta=x^2/a^2+y^2/b^2$$

Using the formula

$$sin^2x+cos^2x=1$$

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

This is the equation of ellipse.

Area of ellipse

=$$4\int_{0}^{a}\frac{b}{a}\sqrt{a^2-x^2}dx$$

When x=0,$$\theta=\pi/2$$

When x=a, $$\theta=0$$

Using the formula

Area of ellipse

=$$\frac{4b}{a}\int_{\pi/2}^{0}\sqrt{a^2-a^2cos^2\theta}(-asin\theta)d\theta$$

Area of ellipse=$$-4ba\int_{\pi/2}^{0}\sqrt{1-cos^2\theta}(sin\theta)d\theta$$

Area of ellipse=$$-4ba\int_{\pi/2}^{0} sin^2\theta d\theta$$

Area of ellipse=$$-2ba\int_{\pi/2}^{0}(2sin^2\theta)d\theta$$

Area of ellipse=$$-2ba\int_{\pi/2}^{0}(1-cos2\theta)d\theta$$

Using the formula

$$1-cos2\theta=2sin^2\theta$$

Area of ellipse=$$-2ba[\theta-1/2sin(2\theta)]^{0}_{\pi/2}$$

Area of ellipse$$=-2ba(-\pi/2-0)$$

Area of ellipse=$$\pi ab$$