# Parametric Equations Finding Area Absolute Value or?

So I am given the following parametric equations.

$$y=bsin(\theta)$$ and $$x=acos(\theta)$$

When I do the following I get a negative area. $$\int_0^{2\pi}b\sin\theta\frac{d}{d\theta}\left(a\cos\theta\right)d\theta$$

I looked on slader and they were doing the following integral:

$$\int_{\frac{\pi}{2}}^0b\sin\theta\left(\frac{d}{d\theta}\left(a\cos\theta\right)\right)d\theta$$

Can someone please explain to me why they did this? Also is taking the absolute value appropriate in cases like this?

• What's the context of the problem? – Patrick Jankowski Nov 12 '18 at 5:27
• To find the area enclosed by the parametric equations. – M.M Nov 12 '18 at 5:28
• What's the domain of $t$? and is the curve oriented anticlockwise or clockwise? – Patrick Jankowski Nov 12 '18 at 5:28
• I am not told the orientation of the curve. the domain is from 0<=theta<=2pi – M.M Nov 12 '18 at 5:29
• Would you be able to write out the question fully? or post a screenshot of it? – Patrick Jankowski Nov 12 '18 at 5:30

The formula for the are under a positive graph and the x-axis is defined to be $$\int _a ^b f(x)dx$$ Where $$a and $$f(x)\ge 0$$
Here $$dx$$ is positive to make the integral positive.
We have to be sure that in our parametric integral for the area, $$\int b \sin \theta d(a\cos \theta)$$ we have our $$d(a\cos (\theta)\ge 0$$ to get a positive answer.
• but they also reduced the domain for $\theta$ to $\large \frac{\pi}{2} \le \theta \le 0$ – Patrick Jankowski Nov 12 '18 at 6:04