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?

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

That is why they swapped the limits of integration.

  • $\begingroup$ but they also reduced the domain for $\theta$ to $\large \frac{\pi}{2} \le \theta \le 0$ $\endgroup$ – Patrick Jankowski Nov 12 '18 at 6:04

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