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I am not sure how to work this one out.

I am suppose to find the area of this parametric equation.

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

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

I set up the equation in the memorized formula.

$$\int_0^{2\pi} \sqrt{1 + \left(\frac{b\cos\theta}{-a\sin\theta}\right)^2}d\theta$$

$$\int_0^{2\pi} \sqrt{1 + \frac{b^2\cos^2\theta}{a^2\sin^2\theta}}d\theta$$

$$\int_0^{2\pi} \sqrt{1 + \frac{b^2}{a^2}\cdot \csc^2\theta}d\theta$$

From here I am at a loss of what to do, I tried some trig subsitution but I do not think that will work and it only seems to complicate the problem.

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    The formula that you used is for the length of the boundary curve, not for the area. If you want to work out the area, maybe you can compute the area that's added as theta changes by a small amount dtheta. The extra area is approximately a triangle. If you work out the area of the triangle and add these up, you should be in business.2012-06-25
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    Please cancel the (differential-equations) tag for this question as this question is not the business of ODE.2012-06-26

4 Answers 4